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https://www.learnmathsonline.org/cbse-class-4-maths/a-trip-to-bhopal-class-4/
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A trip to Bhopal – Class 4.
MODEL QUESTIONS FOR CBSE CLASS 4 MATHEMATICS
A Trip to Bhopal – Chapter 3
1. There are a total of 210 children going to trip. If one bus can take 50 children and if they get 4 buses.
a) How many children will get seats?
b) How many children left without seats?
2. There are a total of 210 children going to trip. Each minibus can take 35 students. How many minibuses are needed?
3. The length of a bridge is 756.82 metres. Can we say that it is more than half a kilometre?
4. Each bus takes about 15 minutes to refill fuel and there are 2 buses to be refilled. So they stop there for about ——– minutes, which means they are late by about ——– minutes?
5. If the number of deer is 117 and number of bison are 37. How many more deer are there than bison?
6. There are a total of 210 children going to a trip. Each child is to be given 1 orange, 1 banana and 5 biscuits. All the children take oranges and biscuits but 38 children do not take bananas. How many oranges, biscuits and bananas are distributed?
7. I gave 4 toffees each to 4 of my friends and 3 toffees are left with me. How many toffees did I have?
8. What are the numbers can you make using 3, 5 and 7?
9. Think of a number which can be divided by 2, 3 and 5 and comes between 25 and 50?
10. A small ant climbs 3 cm in 1 minute but slips down 2 cm. How much time will it take to climb 2 cm?
11. Ticket price for Motorboat is Rs 25 per 20 minutes and boat with oars is Rs 15 per 45 minutes. Indra and Bhanu first went in the Motor boat and then took the oar boat.
a) How much did they pay for both the boats?
b) How much time did they get for both rides?
12. Ticket price for Double decker boat is Rs 30 per 45 minutes. One group of children went for the double decker trip. They paid Rs 450 in total. How many children went for the double decker trip?
1. a) Total number of seats in 4 buses = Number of seats x 4 = 50 x 4 = 200
Hence 200 children will get seats.
b) Total number of students = 210
Number of students who got seats = 200
Children remaining = 210 – 200 = 10
Therefore 10 children are left without seats.
2. Total number of children = 210
Number of children that one minibus can take = 35
Minibus needed = 210/35 = 6.
Therefore, 6 mini buses are needed.
3. Yes, because we know that 1 km = 1000 m.
So half a kilometre = 500m
Since 756.82 are greater than 500, 756.82 is more than half a kilometre.
4. Time taken to refill one bus = 15 minutes.
Time taken to refill two buses = 15 x 2 = 30 minutes
So they are late by about 30 minutes.
5. Number of deer = 117
Number of bison = 37
Number of deer more than bison = 117 – 37 = 80.
6. Number of oranges distributed to 210 children = 210 x 1 = 210
Number of children who took bananas = 210 – 38 = 172
Number of bananas distributed to 210 children = 210 x 5 = 1050
Thus 210 oranges, 172 bananas and 1050 biscuits are distributed.
7. Number of toffees given to 4 friends = 4 x 4 = 16
Number of toffees left = 3
Total number of toffees = 16 + 3 = 19
8. 357, 537, 375, 573, 735, 753
9. The numbers that are divisible by 2 are 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48.
The numbers that are divisible by 3 are 27, 30, 33, 36, 39, 42, 45, 48.
The numbers that are divisible by 5 are 30, 35, 40, 45.
Common number is 30.
So the smallest number that can be divided by 2, 3 and 5 is 30.
10. Ant climbs = 3 cm in 1 minute
Ant slips down = 2 cm in minute
Distance covered by ant in 1 minute = 3 – 2 = 1 cm
So ant will take 2 minutes to climb 2cm.
11. a) Money paid by Indra = 25
Money paid by Bhanu = 25
Total = 25 + 25 = 50 rupees for Moto boat.
Money paid for oar boat = 15 + 15 = 30 rupees
Total money paid = 50 + 30 = 80 rupees.
b) Total time for both the rides = 20 + 45 = 65 minutes.
12. Money spent = Rs 450
Price paid for the double decker boat = Rs 30
Number of children went for the double decker trip = 450/30 = 15 children.
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# Depth First Search and Sudoku
Sudoku is a logic-based puzzle game that involves filling a 9x9 grid with numbers from 1 to 9. The grid is divided into 9 smaller 3x3 grids, and some of the cells are pre-filled with numbers. The objective is to fill in the empty cells with numbers in such a way that each row, column, and 3x3 grid contains all the numbers from 1 to 9 without repetition. Sudoku puzzles can vary in difficulty, with some requiring simple logic and deduction, while others may require more advanced techniques and strategies. The game has gained popularity around the world and is often featured in newspapers, magazines, and puzzle books.
Creating the game of sudoku is one challenge but to have a complete game we will need to provide the solution to each puzzle as well. We could manually do the solutions but this would limit the amount of puzzles and increase the complexity. So we need to be able to automatically solve the puzzles using code. One way that we can use code to solve a sudoku puzzle is with the Depth First Search algorithm. Depth First Search (DFS) is a graph traversal algorithm that explores as far as possible along each branch before backtracking. It starts from a given node in a graph and visits all its adjacent nodes first, filling in valid numbers as it goes. Once no valid numbers can be filled in it then visits the last correct node and changes its number, once changed it continues this process recursively until it reaches a dead end or the goal node, which in this case is the final block filled in with a valid number. DFS maintains a stack to keep track of the visited nodes and the order of their exploration. It uses the stack to backtrack and explore the unvisited branches of the graph until all nodes are visited or the goal node is found. DFS can be used to solve a variety of problems, including finding paths in a maze, determining connected components in a graph, and detecting cycles in a graph. However, DFS may not find the shortest path or optimal solution, as it explores each branch as far as possible before backtracking.
The puzzles are generated by placing a few random numbers which abide by the game rules and then using the Depth First Search algorithm to complete the puzzle, thus creating a valid puzzle. Numbers are then removed from the puzzle at random based on the difficulty level. This method can create millions of valid puzzles with relatively low complexity. The downside to this method is gauging the difficulty of each puzzle.
Below is a game of sudoku. Challenge yourself to complete a puzzle!
How to play:
• Click an empty square
• Type a number on your keyboard to enter it - Type the same number to remove it.
• All rows, columns and 3x3 squares should contain the numbers 1-9 with no duplicates
• If a number appears more than once in a row, colomn or 3x3 square it will be highlighted in red
• You can save a puzzle for later by saving the 'Seed' and entering it in the text box before pressing 'Generate'
Seed:
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# Thread: Introductory linear algebra questions
1. ## Introductory linear algebra questions
Hi guys, having some issues with some basic introductory linear algebra questions, any help much appreciated! - 1) Find the dimension and a basis of the following vector spaces V over the given field K.
(i) V is the set of all vectors (a,b,y) in R^3 with a+2b-2y=0 with K=R.
My answer: Here I'm completely unsure, rearranging the equation for each variable we get the set of vectors (2y-2b,(2y-a)/2, (a+2b)/2). So does this have dimension 3? Clearly any vector can be constructed by (2y-2b)(1,0,0)+((2y-a)/2)(0,1,0)+((a+2b)/2)(0,0,1) and of course (1,0,0),(0,1,0),(0,0,1) are linearly independant. Is this correct?
(ii) V set of vectors (a,b,y) in R^3 with a-b=-y and a+2b=y with K=R.
My answer: Again unsure what to do with the supplied equations. Setting both equations equal to a, then combining them and solving for y gives y=(3/2)b so we have set of vectors (a,b,(3/2)b). Now here I get really confused. Again any vector can be constructed by a(1,0,0)+b(0,1,0)+(3/2)b(0,0,1) and this is the standard basis whose components we know to be linearly independant (or it wouldn't be a basis) which would imply that the vector space has dimension 3. However the any vector can also be constructed by a(1,0,0)+b(0,1,(3/2)) (which are clearly LI) which would imply a dimension of two. Here I'm really confused as the answers are contradictory. I believe the correct dimension to be 2 since the vector space on visualisation is I believe a plane, and so one would expect a dimension of 2 but why do I get two different answers? What am I doing wrong?
This can be generalised to me being unsure of how to approach a situation where I have a set of vectors (a,b,b). This can be spanned in two ways: (1) a(1,0,0)+b(0,1,0)+b(0,0,1) implying a dimension of 3 or (2) a(1,0,0)+b(0,1,1) implying a dimension of 2. In both cases the set of spanning vectors are linearly independant. What gives?
(iii) V=C^2, K=R.
My answer: dimension 4, basis (1,0), (0,i), (0,1), (0,i). Pretty basic textbook example here so no problems (i hope). If K=C then the space would have dimension 2 and a basis (1,0), (0,1), correct?
(iv) V is set of all polynomials over R with degree at most n, in which the sum of the coefficients is 0, K=R.
My answer: So we have a_0+a_1x+a_2x^2+...+a_nx^n with a_0+a_1+a_2+...+a_n=0. Not really sure how the condition of the coefficients restricts things. Normally the set of all polynomials over R with degree at most n would have dimension n+1 and (standard) basis 1,x,x^2,...,x^n. Can someone give me a hint on how to start this one?
(v) K=R and and V is the set of functions from R to R which are solutions of the differential equation d^2f/dx^2-5f=0.
My answer: Solving the (second order) differential equation via standard methods gives a general solution f(x)=c_1e^(root(5)x)+c_2e^(-root(5)x) so does this just have dimension 2 and basis e^(root(5)x), e^(-root(5)x)?
Thanks in advance for any help!
2. ## Re: Introductory linear algebra questions
So does this have dimension 3?
No. y = (a+2b)/2. There is no need to write the other elements of the vector, because that doesn't give you anything. Instead what you have is (a,b,[a+2b]/2), where a and b are real numbers. It's not too hard to show that the basis of this space is (1,0,1/2), (0,1,1).
Now here I get really confused.
You defined a to be one of b-y or y-2b. As you have defined it, a is not a "free variable".
What am I doing wrong?
You have y=b-a and y=a+2b. This is only possible if b=-2a. Then y = -3a. Then the set of vectors is in fact those of the form (a,-2a,-3a), where a is a real number. The dimension should be obvious here.
dimension 4, basis (1,0), (0,i), (0,1), (i,0).
Not really sure how the condition of the coefficients restricts things.
If you know all but one of the coefficients, then you know ALL of the coefficients. Suppose we are in p^2, with the condition on the coefficients. Then a vector is ax^2+ bx+c. But a+b+c= 0 implies that c = -a-b. Then (in vector form) what you really have is (a,0,-a) + (0,b,-b). It should be clear that the basis vectors are (1,0,-1) and (0,1,-1) - with dimension 2.
3. ## Re: Introductory linear algebra questions
Originally Posted by Nemesis10192
Hi guys, having some issues with some basic introductory linear algebra questions, any help much appreciated! - 1) Find the dimension and a basis of the following vector spaces V over the given field K.
(i) V is the set of all vectors (a,b,y) in R^3 with a+2b-2y=0 with K=R.
In addition to ANDS! "y= (a+ 2b)/2" you could also write a= 2y- 2b so that (a, b, y)= (2y- 2b, b, y)= (2y, 0, y)+ (-2b, b, 0)= y(2, 0, 1)+ b(-2, 1, 0) which should make it clear what a basis and the dimension are.
(ii) V set of vectors (a,b,y) in R^3 with a-b=-y and a+2b=y with K=R.
a- b= -y is the same as y= b- a so (a, b, y)= (a, b, b- a)= (a, 0, -a)+ (0, b, b)= a(1, 0, -1)+ b(0, 1, 1).
This can be generalised to me being unsure of how to approach a situation where I have a set of vectors (a,b,b). This can be spanned in two ways: (1) a(1,0,0)+b(0,1,0)+b(0,0,1) implying a dimension of 3
NO. Those are not two different "b"s so you cannot write them separately.
or (2) a(1,0,0)+b(0,1,1) implying a dimension of 2. In both cases the set of spanning vectors are linearly independant. What gives?
This is correct because it has just the one value of "b".
(iii) V=C^2, K=R.
My answer: dimension 4, basis (1,0), (0,i), (0,1), (0,i). Pretty basic textbook example here so no problems (i hope). If K=C then the space would have dimension 2 and a basis (1,0), (0,1), correct?
Yes, $\displaystyle C^2$, as a vector space over the real numbers has dimension 4, as a vector space over the complex numbers, dimension 2..
(iv) V is set of all polynomials over R with degree at most n, in which the sum of the coefficients is 0, K=R.
My answer: So we have a_0+a_1x+a_2x^2+...+a_nx^n with a_0+a_1+a_2+...+a_n=0. Not really sure how the condition of the coefficients restricts things. Normally the set of all polynomials over R with degree at most n would have dimension n+1 and (standard) basis 1,x,x^2,...,x^n. Can someone give me a hint on how to start this one?
The "set of all polynomials over R with degree at most n" is a vector space of dimension n, but the condition that $\displaystyle a_0+a_1+a_2+...+a_n=0$ allows you to write one as a linear combination of the others, $\displaystyle a_0= -(a_1+a_2+...+a_n)$, so that you can replace that one with the others. That means you are left with n-1 "independent" numbers so any such polynomial can be written as a linear combination of those n-1 terms.
(v) K=R and and V is the set of functions from R to R which are solutions of the differential equation d^2f/dx^2-5f=0.
My answer: Solving the (second order) differential equation via standard methods gives a general solution f(x)=c_1e^(root(5)x)+c_2e^(-root(5)x) so does this just have dimension 2 and basis e^(root(5)x), e^(-root(5)x)?
Thanks in advance for any help!
Yes, that's correct. More generally, the set of all solutions to any nth order linear, homogeneous, differential equation is a vector space of dimension n.
4. ## Re: Introductory linear algebra questions
Many thanks for your replies guys, they cleared up things immensely for me. That said, I would like to check my answers to some new questions since and have decided its probably better just to continue within this thread. Hopefully you guys will have the time to help me again . Read up on how to use basic latex+ got an image of the problem sheet so hopefully my post will be a bit more readable this time! Here is the problem sheet(insert picture wasn't working-I press upload and nothing happens):
Nemesis10192's Library | Photobucket
(1) (i) yes (ii) yes (iii) yes (iv) no- Rewriting using the supplied equation we have $\displaystyle (\alpha,\beta,\frac{\alpha+\beta-1}{7})$ thus testing that scalar multiplication and vector addition returns another vector in the subspace we have $\displaystyle a(\alpha_{1}, \beta_{1}, \frac{\alpha_{1}+\beta_{1}-1}{7}) + b(\alpha_{2}, \beta_{2}, \frac{\alpha_{2}+\beta_{2}-1}{7})=(a\alpha_{1}+b\alpha_{2}, a\beta_{1}+b\beta_{2},\frac{(a\alpha_{1}+b\alpha_{ 2})+(a\beta_{1}+(b\beta_{2})-(a+b)}{7})$ which is of the form $\displaystyle (\alpha, \beta, \frac{\alpha+\beta-(a+b)}{7})$ and so by the (a+b) term at the end instead of just 1 we see that the answer to (iv) is no. Are these correct?
(2) Not too bothered about these, fairly routine- dimension= number of vectors in the basis, can find basis by sifting and we may stop once we have 4 vectors anyhow since vectors with 4 components cannot possibly span anything with more than dim=4, correct? About (ii) though- I'm unsure what to make of the condition about the finite field, does it literally just restrict the components of each vector to being 0 or 1, or does it affect the answer in some other way?
(3) Would really like an assessment of how good my proofs are for the following question. I'm new to proof writing so don't really know what's acceptable/how nitpicky I need to be etc. If for any part there is a more succinct way of proving it, please do share .
(i) Well this is very obvious, so much so that I'm not really sure how to prove it. My best attempt is something along the lines of:
Since $\displaystyle w_{1},...,w_{m}$ do not span W we can pick a $\displaystyle w_{m+1}$ s.t. it is not a linear combination of $\displaystyle w_{1},...,w_{m}$ but since $\displaystyle w_{1},...,w_{m}$ are also linearly independent we have $\displaystyle a_{1}w_{1}+a_{2}w_{2}+...+a_{m}w_{m}+a_{m+1}w_{m+1 }=0 \Rightarrow a_{1}=a_{2}=a_{m}=a_{m+1}=0$ which is the definition of linearly independence hence $\displaystyle w_{1},...,w_{m+1}$ are linearly independent.
(ii) Proof of finiteness of dimension of W: Since V is finite dimensional lets call its dimension n. Assume W is infinite dimensional, then we can find vectors $\displaystyle w_{1},w_{2},...,w_{m} m>n$ that are linearly independent since an infinite dimensional space has infinitely many linearly independent vectors in it. But since W is a subgroup of V these m linearly independent vectors must also be in V which contradicts dim(v)=n since dimension of a vector space is defined to be the number of linearly independent vectors in any basis for it. Therefore W is finite-dimensional.
Proof that $\displaystyle dim(W)\leq dim(V):$$\displaystyle Let w_{1},...,w_{m}$ be a basis for W. Since these vectors form a basis for W, they are linearly independent, and since W is a subgroup of V then $\displaystyle w_{1},...,w_{n} n\geq m$ is a linearly independent set of vectors in V. Therefore we can extend the basis for W into a basis for V; $\displaystyle w_{1},...,w_{m},v_{1}...,v_{k} k\geq 0$. Hence $\displaystyle dim(V)=m+k \geq m=dim(W).$
(iii) Since V is finite-dimensional, we know by (ii) that $\displaystyle W_{1}$ and $\displaystyle W_{2}$ are finite-dimensional. Let $\displaystyle w_{1},...,w_{m}$ be a basis for $\displaystyle W_{1}.$Then since $\displaystyle W_{1}$ is a subgroup of$\displaystyle W_{2}, w_{1},...,w_{n}$ is a set of vectors in $\displaystyle W_{2}.$By definition of basis $\displaystyle w_{1},...,w_{m}$ is a linearly independent set and so can be extended to a basis for $\displaystyle W_{2}; w_{1},...,w_{m},v_{1},...v_{k} k \geq 0$. Since $\displaystyle dim(W_{1})=dim(W_{2})$we have $\displaystyle m=dim(W_{1})=dim(W_{2})=m+k \Rightarrow m=m+k \Rightarrow k=0$ and so by the way we have set up our bases we see that $\displaystyle W_{1}$ and $\displaystyle W_{2}$ have the same basis $\displaystyle w_{1},...,w_{m}$and so have the same span $\displaystyle \Rightarrow W_{1}=W_{2}.$
(4) W is again finite dimensional by (ii). Let a basis for W be $\displaystyle w_{1},...,w_{m}$ then since W is a subgroup of V, $\displaystyle w_{1},...,w_{n}$ is a set of vectors in V. Since $\displaystyle w_{1},...,w_{m}$ is a basis it is a set of linearly independent vectors and so can be extended to a basis for V; $\displaystyle w_{1},...,w_{m},x_{1},...,x_{k} k \geq 0$. Since the aforementioned set is a basis, its members are linearly independent, therefore $\displaystyle x_{1},...,x_{k}$ are linearly independent. Define a new subspace X of V generated by the basis $\displaystyle x_{1},...,x_{k} k\geq 0.$ Then V=W+X and $\displaystyle W\bigcap X=0$ and so we see W has a complementary subgroup in V.
Critique of the proofs very much wanted!
(5) (i) basis of image (1,0,1,0),(0,0,1,1),(1,1,0,0),(0,1,0,1); Rank=4. By rank-nullity we therefore have nullity=0 and a basis of the kernel=$\displaystyle 0_{V}$?
(ii) So in general we have the mapping of $\displaystyle a+bx+cx^2+dx^3+ex^4+fx^5$ to $\displaystyle 24e+120fx$. Therefore a basis of the image is 1,x and rank=2? Therefore nullity=4 and a basis of the kernel is $\displaystyle 1,x,x^2,x^3$? Bit confused about this one, as well as the concept of a kernel.
(6) (i) Stretch scale factor 2 and rotation 180 degrees about origin.
(ii) Projection onto the y axis.
(iii) reflection in y=x.
(iv) Don't know how to describe the geometrically- x coordinate maps to x+y and y coordinate maps to y-x then a scale factor of$\displaystyle \frac{1}{\sqrt{2}}$is applied to the whole thing but I have no idea how to describe this succinctly?
(7) No idea here, can someone give me some starting hints please?
Wow, that took depressingly long to type out! Hope I become faster with familiarity . Thanks in advance for any help!
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you are taking the following test:
Grade: 6, Subject: Math, Topic: Percents Percent, calculating simple interest
Question 1: 24 is 6% of what number? 240 400 600 260
Question 2: Susan bought one color printer for \$95 and two color refill cartridges for \$42.50 each. If the sales tax were 7.25%, how much tax would she pay for purchase? \$12.75 \$12.70 \$13.05 \$13.75
Question 3: A computer that usually sells for \$856 goes on sale for \$630. What is the percent decrease, to the nearest tenth of a percent? 73.6% 26.4% 135.5% 26.9%
Question 4: Marcus borrowed \$3000 for 5 years at simple interest to pay for his car. If Marcus repaid a total of \$3,750.00, at what interest rate did he borrow the money? 2% 15% 8.5% 5%
Question 5: Write 400% as a decimal. 0.4 0.04 4.0 1.04
Question 6: Find the interest. Principal = 9,000 Rate = 6.5% Time in years = 8 1,7307.69 \$4,680 \$40,680 \$4,680.85
Question 7: Mr. Robinson opened a saving account that earns 3.5% annual interest. He wants to earn at least \$315 in interest after 3 years. How much money should he invest in order to earn \$315 in interest? \$30 \$270 \$3,500 \$3,000
Question 8: What is 28.5% of 850? 242.25 42.25 2,422.5 202.25
Question 9: 16 is 20% of what number? 80 40 32 60
Question 10: Sarah now earns \$12.50 per hour. This is 125% of what she earned last year. How much did she earn per hour last year? \$12 \$8.75 \$10 \$15
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# How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?
I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$, where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation \begin{align} & s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\ & \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right), \end{align} but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
• Is that just $(H_n)^2$ or $H_n^{(2)}$? – Ali May 29 '13 at 4:22
• Looking at the definition of $H_n$, lack of $\,{}^{(\phantom2)}$ around ${}^2$ and the values of partial sums, I would assume $(H_n)^2$. – Vladimir Reshetnikov May 29 '13 at 4:38
• The $s$ sequences make a real vector space of dimension 4. So, if you manage to find the sequences with $(s_1, s_2,s_3,s_4) = (1,0,0,0),$ then $(s_1, s_2,s_3,s_4) = (0,1,0,0),$ then $(s_1, s_2,s_3,s_4) = (0,0,1,0),$ then $(s_1, s_2,s_3,s_4) = (0,0,0,1),$ you are set. – Will Jagy May 29 '13 at 4:50
• – math110 May 29 '13 at 5:22
• @math110: There may be some useful ideas there, but $H_n^2\ne H_n^{(2)}$, so this isn’t $A(2,1)$. – Brian M. Scott May 29 '13 at 5:24
Write down the function $$g(z) = \sum_{n\geq1} \frac{z^n}{n}H_n^2,$$ so that $S=g(-1)$ and $g$ can be reduced to $$zg'(z) = \sum_{n\geq1} z^n H_n^2 = h(z).$$
Now, using $H_n = H_{n-1} + \frac1n$ ($n\geq2$), we can get a closed form for $h(z)$: $$h(z) = z + \sum_{n\geq2}\frac{z^n}{n^2} + \sum_{n\geq 2}z^n H_{n-1}^2 + \sum_{n\geq 2} 2\frac{z^n}{n}H_{n-1}.$$ Now, the first and third sums Mathematica can evaluate itself in closed form (the third one evaluates to the function $p(z)$ below, the first one is $\text{Li}_2(z)-z$), and the middle sum is $z h(z)$.
Substituting this into the expression for $g(z)$, we get $$g(z) = \int \frac{\text{Li}_2(z) + p(z)}{z(1-z)}\,dz,$$ $$p(z) = -\frac{\pi^2}{3} + 2\log^2(1-z)-2\log(1-z)\log(z)+2\text{Li}_2((1-z)^{-1}) - 2\text{Li}_2(z).$$ Mathematica can also evaluate this integral, giving (up to a constant of integration) \begin{align} g(z) &= \frac{1}{3} \left(-2 \log(1-z^3+3 \log(1-z)^2 \log(-z)+\log(-1+z)^2 (\log(-1+z)+3 \log(-z) \right. \\ & \hspace{5mm} \left. -3 \log(z))+\pi ^2 (\log(-z)-2 \log(z))+\log(1-z) \left(\pi^2 - 3 \log(-1+z)^2 \right. \right.\\ & \hspace{5mm} \left.\left. +6 (\log(-1+z)-\log(-z)) \log(z)\right)-6 (\log(-1+z)-\log(z)) \left(\text{Li}_{2}\left(\frac{1}{1-z}\right)-\text{Li}_{2}(z)\right) \right.\\ & \hspace{10mm} \left. -3 \log(1-z) \text{Li}_{2}(z)+3 \text{Li}_{3}(z)\right). \end{align} The constant of integration is fixed by requiring $g(0)=0$.
Some care needs to be taken, because the function is multi-valued, when evaluating $g(-1)$. The answer is $$\frac{1}{12}(\pi^2\log2-4(\log 2)^3-9\zeta(3)).$$
let $$y=\sum_{n=1}^{\infty}H^2_{n}x^n$$
then we have $$y=x+xy+\ln^2{(1-x)}+\int_{0}^{x}\dfrac{\ln{(1-t)}}{t}dt$$
so $$y=\dfrac{\ln^2{(1-x)}}{1-x}+\sum_{n=1}^{\infty}\left(1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}\right)x^n$$
we have $$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
replace $$x$$ with $$-x$$, divide both sides by $$x$$ then integrate w.r.t $$x$$ from $$0$$ to $$1$$ , we get: \begin{align*} S_1&=\sum_{n=1}^{\infty}(-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ dx=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\frac{(-1)^n}n=\underbrace{\int_0^1\frac{\ln^2(1+x)}{x(1+x)}\ dx}_{x=\frac{1-y}{y}}\\ &=\int_{1/2}^1 \frac{\ln^2x}{1-x}\ dx=\sum_{n=1}^{\infty}\int_{1/2}^1x^{n-1}\ln^2x\ dx=\sum_{n=1}^{\infty}\left(\frac{2}{n^3}-\frac{2}{2^n n^3}-\frac{2\ln2}{2^n n^2}-\frac{\ln^22}{2^n n}\right)\\ &=2\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-2\ln2\operatorname{Li_2}\left(\frac12\right)-\ln^32 \end{align*}
Now using the identity:
$$\displaystyle \int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac1n\left(H_n^2+H_n^{(2)}\right)$$
multiply both sides by $$(-1)^n$$ then sum both sides w.r.t $$n$$ from $$1$$ to $$\infty$$, we get \begin{align*} S_2&=\sum_{n=1}^{\infty}\left(H_n^2+H_n^{(2)}\right)\frac{(-1)^n}{n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^{\infty}(-x)^n\ dx=\underbrace{-\int_0^1\frac{\ln^2(1-x)}{1+x}\ dx}_{x=1-y}\\ &=-\int_0^1\frac{\ln^2(x)}{2-x}=-\sum_{n=1}^{\infty}\frac1{2^n}\int_0^1 x^{n-1}\ln^2x\ dx=-2\sum_{n=1}^{\infty}\frac1{2^n n^3}=-2\operatorname{Li_3}\left(\frac12\right) \end{align*} we are now ready to calcualate our sum: \begin{align*} \frac{S_1+S_2}{2}=\sum\frac{(-1)^n H_n^2}{n}&=\zeta(3)-2\operatorname{Li_3}\left(\frac12\right)-\ln2\operatorname{Li_2}\left(\frac12\right)-\frac12\ln^32\\ &=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32 \end{align*}
and as a bonus : \begin{align*} \frac{S_2-S_1}{2}=\sum\frac{(-1)^n H_n^{(2)}}{n}&=\ln2\operatorname{Li_2}\left(\frac12\right)-\zeta(3)+\frac12\ln^32\\ &=\frac12\ln2\zeta(2)-\zeta(3) \end{align*}
where the results of $$\operatorname{Li_3}\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ and $$\operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ were used in the calculations.
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# parasys.net
Home > Error Propagation > Error Propagation Through Subtraction
# Error Propagation Through Subtraction
## Contents
The size of the error in trigonometric functions depends not only on the size of the error in the angle, but also on the size of the angle. If you are converting between unit systems, then you are probably multiplying your value by a constant. Article type topic Tags Upper Division Vet4 © Copyright 2016 Chemistry LibreTexts Powered by MindTouch ERROR ANALYSIS: 1) How errors add: Independent and correlated errors affect the resultant error in Why can this happen? More about the author
All the rules that involve two or more variables assume that those variables have been measured independently; they shouldn't be applied when the two variables have been calculated from the same When two quantities are multiplied, their relative determinate errors add. This result is the same whether the errors are determinate or indeterminate, since no negative terms appeared in the determinate error equation. (2) A quantity Q is calculated from the law: Summarizing: Sum and difference rule.
## Error Propagation Addition And Subtraction
The student might design an experiment to verify this relation, and to determine the value of g, by measuring the time of fall of a body over a measured distance. It's a good idea to derive them first, even before you decide whether the errors are determinate, indeterminate, or both. We quote the result in standard form: Q = 0.340 ± 0.006.
Error Propagation Contents: Addition of measured quantities Multiplication of measured quantities Multiplication with a constant Polynomial functions General functions Very often we are facing the situation that we need to measure Two numbers with uncertainties can not provide an answer with absolute certainty! The resultant absolute error also is multiplied or divided. Error Propagation Calculator For example, if some number A has a positive uncertainty and some other number B has a negative uncertainty, then simply adding the uncertainties of A and B together could give
For example, to convert a length from meters to centimeters, you multiply by exactly 100, so a length of an exercise track that's measured as 150 ± 1 meters can also Uncertainty Subtraction In summary, maximum indeterminate errors propagate according to the following rules: Addition and subtraction rule. What is the error in R? https://www.lhup.edu/~dsimanek/scenario/errorman/propagat.htm When we are only concerned with limits of error (or maximum error) we assume a "worst-case" combination of signs.
The results of each instrument are given as: a, b, c, d... (For simplification purposes, only the variables a, b, and c will be used throughout this derivation). Error Propagation Square Root Try all other combinations of the plus and minus signs. (3.3) The mathematical operation of taking a difference of two data quantities will often give very much larger fractional error in These rules only apply when combining independent errors, that is, individual measurements whose errors have size and sign independent of each other. The derivative with respect to x is dv/dx = 1/t.
## Uncertainty Subtraction
General function of multivariables For a function q which depends on variables x, y, and z, the uncertainty can be found by the square root of the squared sums of the http://chem.libretexts.org/Core/Analytical_Chemistry/Quantifying_Nature/Significant_Digits/Propagation_of_Error The derivative, dv/dt = -x/t2. Error Propagation Addition And Subtraction How precise is this half-life value? Error Propagation Formula Physics A pharmacokinetic regression analysis might produce the result that ke = 0.1633 ± 0.01644 (ke has units of "per hour").
This is an example of correlated error (or non-independent error) since the error in L and W are the same. The error in L is correlated with that of in W. my review here Now we are ready to use calculus to obtain an unknown uncertainty of another variable. What is the error in the sine of this angle? Then it works just like the "add the squares" rule for addition and subtraction. Error Propagation Average
Also, notice that the units of the uncertainty calculation match the units of the answer. The fractional indeterminate error in Q is then 0.028 + 0.0094 = 0.122, or 12.2%. First you calculate the relative SE of the ke value as SE(ke )/ke, which is 0.01644/0.1633 = 0.1007, or about 10 percent. http://parasys.net/error-propagation/error-propagation-subtraction-constant.php So if the angle is one half degree too large the sine becomes 0.008 larger, and if it were half a degree too small the sine becomes 0.008 smaller. (The change
The fractional determinate error in Q is 0.028 - 0.0094 = 0.0186, which is 1.86%. Error Propagation Chemistry This, however, is a minor correction, of little importance in our work in this course. In this example, the 1.72 cm/s is rounded to 1.7 cm/s.
## etc.
This step should only be done after the determinate error equation, Eq. 3-6 or 3-7, has been fully derived in standard form. which rounds to 0.001. In lab, graphs are often used where LoggerPro software calculates uncertainties in slope and intercept values for you. Error Propagation Inverse If you're measuring the height of a skyscraper, the ratio will be very low.
Principles of Instrumental Analysis; 6th Ed., Thomson Brooks/Cole: Belmont, 2007. All rules that we have stated above are actually special cases of this last rule. Answer: we can calculate the time as (g = 9.81 m/s2 is assumed to be known exactly) t = - v / g = 3.8 m/s / 9.81 m/s2 = 0.387 navigate to this website Please try the request again.
However, we want to consider the ratio of the uncertainty to the measured number itself. However, in complicated scenarios, they may differ because of: unsuspected covariances errors in which reported value of a measurement is altered, rather than the measurements themselves (usually a result of mis-specification Solution: First calculate R without regard for errors: R = (38.2)(12.1) = 462.22 The product rule requires fractional error measure. When the error a is small relative to A and ΔB is small relative to B, then (ΔA)(ΔB) is certainly small relative to AB.
So the result is: Quotient rule. Results are is obtained by mathematical operations on the data, and small changes in any data quantity can affect the value of a result. Multiplication or division, relative error. Addition or subtraction: In this case, the absolute errors obey Pythagorean theorem. If a and b are constants, If there Accounting for significant figures, the final answer would be: ε = 0.013 ± 0.001 L moles-1 cm-1 Example 2 If you are given an equation that relates two different variables and
The fractional error in the denominator is 1.0/106 = 0.0094. Consider a result, R, calculated from the sum of two data quantities A and B. The equation for molar absorptivity is ε = A/(lc). Note that this fraction converges to zero with large n, suggesting that zero error would be obtained only if an infinite number of measurements were averaged!
The fractional error may be assumed to be nearly the same for all of these measurements. What is the uncertainty of the measurement of the volume of blood pass through the artery? Hint: Take the quotient of (A + ΔA) and (B - ΔB) to find the fractional error in A/B. It is a calculus derived statistical calculation designed to combine uncertainties from multiple variables, in order to provide an accurate measurement of uncertainty.
It's easiest to first consider determinate errors, which have explicit sign. When mathematical operations are combined, the rules may be successively applied to each operation. In problems, the uncertainty is usually given as a percent. It can show which error sources dominate, and which are negligible, thereby saving time you might otherwise spend fussing with unimportant considerations.
We conclude that the error in the sum of two quantities is the sum of the errors in those quantities. as follows: The standard deviation equation can be rewritten as the variance ($$\sigma_x^2$$) of $$x$$: $\dfrac{\sum{(dx_i)^2}}{N-1}=\dfrac{\sum{(x_i-\bar{x})^2}}{N-1}=\sigma^2_x\tag{8}$ Rewriting Equation 7 using the statistical relationship created yields the Exact Formula for Propagation of The error in g may be calculated from the previously stated rules of error propagation, if we know the errors in s and t.
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New
New
Year 4
# Solve problems using the relationship between 3 and 9 times table
I can solve problems using the relationship between pairs of 3 and 9 times table facts with the same product.
New
New
Year 4
# Solve problems using the relationship between 3 and 9 times table
I can solve problems using the relationship between pairs of 3 and 9 times table facts with the same product.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. There are three times as many threes as nines.
2. We can say that triple three is equal to nine.
3. Triple means 'three times'.
### Common misconception
Pupils may find it difficult to solve problems or misunderstand questions.
Model clearly how to identify key phrases and words which will identify the operation and equation required for word problems. This could involve drawing bar models or other representations.
### Keywords
• Triple - To triple means to become three times as many or to multiply by three.
By this point fluency in 3 and 9 times tables will support children in solving these problems. Continue to chorally rehearse the 3 and 9 times tables.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
Which calculation does the image represent?
2 × 3
3 × 3
4 × 3
Q2.
Three times as many means …
to multiply.
to double.
Q3.
Match the expressions to the product.
18
21
24
27
36
Q4.
If I know 3 × 4 = 12, then I know 9 × 4=
Q5.
Match the groups of threes to the groups of 9 with the same value.
Correct Answer:6 groups of 3,2 groups of 9
2 groups of 9
Correct Answer:3 × 9,9 groups of 3
9 groups of 3
Correct Answer:3 groups of 12,4 × 9
4 × 9
Correct Answer:3 × 15,5 groups of 9
5 groups of 9
Q6.
If Izzy has collected 30 guinea cards and Andeep has triple the amount, how many cards does he have altogether? 30 × 3 =
## Exit quiz
### 6 Questions
Q1.
Which numbers are a multiple of 3?
5
29
Q2.
Identify the multiples of 9
8
26
35
Q3.
Look at the word problem, which words suggest we must divide? A baker baked 36 cupcakes. He splits them equally into boxes of 9, how many boxes will he need?
baked
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# Calculus
posted by .
Find:
(a) the interval(s) on which f is increasing,
(b) the interval(s) on which f is decreasing,
(c) the open interval(s) on which f is concave up,
(d) the open interval(s) on which f is concave down, and
(e) the X-coordinate(s) of any/all inflection point(s).
f(x)= (x^4) - (4802 x^2) + 9604
• Calculus -
take the first derivative ....
a) f is increasing for all values of x for which f '(x) is positive
f is decreasing for all values of x for which f '(x) is negative
take the second derivative
c) if the second derivative is positive, f is concave up
d) if the second derivative is negative, f is concave down
e) set the second derivative equal to zero and solve for x
f '(x) = 4x^3 - 9604x
= 4x(x^2 - 2401)
= 4x(x-49)(x+49)
What conclusions can you draw from that?
Use what you know about the properties and general shape of y = x^4 + .....
• Calculus -
i still did not understand. if u don't mind could you please explain it again and also the question asks for the intervals and the points. could you give the intervals and points for the the questions above in the questions .. Thanks a bunch :).
## Similar Questions
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Find: (a) the interval(s) on which f is increasing, (b) the interval(s) on which f is decreasing, (c) the open interval(s) on which f is concave up, (d) the open interval(s) on which f is concave down, and (e) the X-coordinate(s) of …
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8. ### Calculus
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https://convertoctopus.com/118-ounces-to-pounds
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## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 118 ounces into pounds we have to multiply 118 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
118 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 118 oz × 0.0625 lb
M(lb) = 7.375 lb
The final result is:
118 oz → 7.375 lb
We conclude that 118 ounces is equivalent to 7.375 pounds:
118 ounces = 7.375 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.13559322033898 × 118 ounces.
Another way is saying that 118 ounces is equal to 1 ÷ 0.13559322033898 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred eighteen ounces is approximately seven point three seven five pounds:
118 oz ≅ 7.375 lb
An alternative is also that one pound is approximately zero point one three six times one hundred eighteen ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
119 ounces 7.438 pounds
120 ounces 7.5 pounds
121 ounces 7.563 pounds
122 ounces 7.625 pounds
123 ounces 7.688 pounds
124 ounces 7.75 pounds
125 ounces 7.813 pounds
126 ounces 7.875 pounds
127 ounces 7.938 pounds
128 ounces 8 pounds
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# Find out the total Area of the shaded figure.
This question was previously asked in
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1. 10920 sq. mt
2. 394 sq. mt
3. 446 sq. mt
4. 204 sq. mt
Option 3 : 446 sq. mt
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## Detailed Solution
Given:
Formula Used:
Area of rectangle = Length × Breadth
Calculation:
One rectangle length = 70 m and breadth = 6 m
Area of rectangle = 70 × 6 = 420 m2
Second rectangle length = 2 m and breadth = 13 m
Area of rectangle = 2 × 13 = 26 m2
According to the question,
Sum of both rectangle = 420 + 26
⇒ 446 m2
∴ The total Area of the shaded figure is 446 sq. mt
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Question: What Is The Prime Factorization Of 171?
What table does 171 come in?
The value can not be more than 200171 Times Table1x1712x3423x5134x68421 more rows
What is the prime factorization of 450?
The prime factorization of 450 is 2 × 3 × 3 × 5 × 5. Written with exponents, the answer is 2 × 32 × 52.
What is the prime factorization of 180?
The prime factorization of 180 is 5 × 2 × 2 × 3 × 3. We can also represent this prime factorization as 5 × 22 × 32…
Is 171 prime or composite?
For 171, the answer is: No, 171 is not a prime number. The list of all positive divisors (i.e., the list of all integers that divide 171) is as follows: 1, 3, 9, 19, 57, 171. For 171 to be a prime number, it would have been required that 171 has only two divisors, i.e., itself and 1.
What is the perfect square of 171?
Q: Is 171 a Perfect Square? A: No, the number 171 is not a perfect square.
Is 91 a prime number Yes or no?
To find whether 91 is a prime number, first, check whether (9+1) is divisible by 3. Since 9+1 is 10 which is not divisible by 3, continue with the following steps. … So, 91 is divisible by 7 other than 1 and itself. Hence, it is not a prime number.
What is the prime factorization of 43?
Prime factorization: 43 is prime. The exponent of prime number 43 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 43 has exactly 2 factors.
What is the HCF of 5 10 and 15?
Factors of 10 = 1, 2, 5 and 10. Therefore, common factor of 15 and 10 = 1 and 5. Highest common factor (H.C.F) of 15 and 10 = 5.
What are the prime factors of 171?
Q: What is the prime factorization of the number 171?The prime factors are: 3 x 3 x 19. or also written as { 3, 3, 19 }Written in exponential form: 32 x 191
What is the prime factorization of 172?
The prime factorization of 172 = 22•43.
What are the factors of 173?
173 and Level 6173 is a prime number.Prime factorization: 173 is prime.The exponent of prime number 173 is 1. Adding 1 to that exponent we get (1 + 1) = 2. … Factors of 173: 1, 173.Factor pairs: 173 = 1 x 173.173 has no square factors that allow its square root to be simplified. √173 ≈ 13.1529.
IS 172 a perfect square?
Is 172 a perfect square number? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 172 is about 13.115. Thus, the square root of 172 is not an integer, and therefore 172 is not a square number.
Is 57 divisible by any number?
For 57, the answer is: No, 57 is not a prime number. The list of all positive divisors (i.e., the list of all integers that divide 57) is as follows: 1, 3, 19, 57.
What are the multiples of 171?
Answer : 171,342,513,684,855,1026,1197,1368,1539,1710,1881,2052,2223,2394,2565,2736,2907,3078,3249,3420,3591,3762,3933,4104,4275,4446,4617,4788,4959,5130,5301,5472,5643,5814,5985,6156,6327,6498,6669,6840,7011,7182,7353,7524,7695,7866,8037,8208,8379, Related Links : What are the factors of 171?
What are the factors of 174?
Factors of 174 are 1, 2, 3, 6, 29, 58, 87.
Is 173 a prime number Why?
For 173, the answer is: yes, 173 is a prime number because it has only two distinct divisors: 1 and itself (173).
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# Math
(a) translate the argument into symbolic form and (b) determine if the argument is valid or invalid. You may compare the argument to a standard form or use a truth table.
If I can get my child to preschool by 9:45AM, then I can take the 9:00AM class.
If I can take the 9AM class, then I can be done by 2PM
If I can get my child to preschool by 8:45AM then I can be done by 2PM
1. 👍 0
2. 👎 0
3. 👁 192
1. p -> q
q -> r
p -> r
valid
1. 👍 0
2. 👎 0
posted by Steve
2. p_>r
1. 👍 0
2. 👎 0
posted by Shazia
3. 20 x 20 = 100
1. 👍 0
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posted by leshem
4. 400
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2. 👎 0
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### Home > AC > Chapter 7 > Lesson 7.2.1 > Problem7-60
7-60.
Dominic simplified an expression using the Distributive Property and got this result: $15x^2-5x$. Can you find a possible expression that he started with?
What are the factors of $15x^2$ and $5x$?
There are a few possible answers.
One answer is: $x\left(15x-5\right)$
Can you think of another answer?
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Home | Lesson Archive | Year 3 Maths Archive | L1: unexpected guests
# Fractions of quantities: L1: unexpected guests
## The Lesson
In this exciting lesson, you will have an opportunity to show your understanding of halves, thirds and quarters, whilst solving a variety of fraction problems that relate to the fascinating clip that you watched at the start of the lesson and the animals that it featured.
Useful prior knowledge:
• To know what is meant by the numerator and the denominator in a unit fraction.
• To be able to recognise a half (1/2), a third (1/3) and a quarter 1/4).
• To be able to calculate unit fractions of small quantities (e.g. 1/2 of 8, 1/4 of 12 etc.).
Did you know?
An elephant’s trunk has more than 40 thousand muscles. The trunks are used for sucking up water, picking up or or touching objects, trumpeting warnings, and for greeting one another. An elephant can also use its trunk as a snorkel when swimming!
## Whiteboard Challenges
Divide the bars and draw counters to solve the problems below.
A herd of 12 elephants stopped at the Zambezi River before walking through the safari lodge to get to the mango tree.
1) At the river, 1/2 of the herd squirted water over their backs to keep themselves cool.
How many elephants squirted water over their backs?
2) At the river, 1/3 of the herd used their trunk like a snorkel.
How many elephants used their trunk like a snorkel?
3) At the river, 1/4 of the elephants in the herd were more than 40 years old.
How many elephants were more than 40 years old?
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# How do you evaluate (1 + 3 * 3)^2 + 8?
Mar 20, 2018
108
#### Explanation:
• Remember that PEMDAS:
• List item
• List item
• Parenthesis
• Exponents
• Multiplication
• Division
• Subtraction
*Remember that steps 3/4 can happen in either order depending on what appear first when reading the equation from left to right. This same rule applies with steps 5/6 as well *
is the necessary order of operations to solve all equations.
• Solving the Equation:
${\left(1 + 3 \cdot 3\right)}^{2} + 8$
- to begin, rewrite the problem, and identify the first step according to PEMDAS
- The first letter of PEMDAS, P stands for parenthesis, which are present
- So, we begin by solving inside the parenthesis
${\left(1 + 3 \cdot 3\right)}^{2} + 8$
• Inside the parenthesis, we restart with PEMDAS: Since there are no parenthesis or exponents INSIDE the parenthesis, we move on to multiplication - 3 x 3 = 9 - so the equation should look like this:
${\left(1 + 9\right)}^{2} + 8$
• To finish the parenthesis, we move down PEMDAS - since there is no more multiplication or division, we can add -
• 1 + 9 = 10 - so the parenthesis are removed because they have been simplified, and the equation should look like this:
${10}^{2} + 8$
• Then, we return to the beginning of PEMDAS - since there are no more parenthesis, we can move on to the exponent that is present.
• ${10}^{2} = 100$ - so, the number 100 can fill the place of ${10}^{2}$ in the equation
$100 + 8$
• Finally, we can move on to the next part of PEMDAS - since there is no more multiplication or division, we can move on addition and subtraction
$100 + 8 = 108$ - so you have your answer!
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Thread: Continuity of a 2 variable function
1. Continuity of a 2 variable function
Let $A=\{ (x,y) \in \mathbb{R}^2 :y \leq x-1 \}$ and let $f(x,y)$ be defined in all $\mathbb{R}^2$ by $f(x,y)= \begin{cases} y^2-(x-1)^2 \hspace{2cm} \text{if (x,y)} \in A \\ 0 \hspace{3 cm} \text{if (x,y) is not in A} \end{cases}$.
1)Analyze the continuity of $f$ in all $\mathbb{R}^2$.
I'll put the other questions if I need help, but I'd like a confirmation of my result for part 1).
My attempt :
First I tried to visualize the function and it was quite hard. I couldn't even do that. I realize that it takes the element $y=x-1$ (a line) from the domain and send it to the image as the line $y=0$ (or curve?). I also realize that it sends all the elements under the curve $y=x-1$ into another area, but I didn't visualize it. Anyway, back to the question, I think the limit exists by intuition.
Here's my work to support this : $f$ is continuous in all $\mathbb{R}^2$ except maybe in $y=x-1$. I'll try to prove it is :
I want to prove that $\lim _{y \to x-1} f(x,y)=0$.
$\forall \varepsilon >0$, $\exists \delta (\varepsilon)$ such that if $|(x,y)-(x,x-1)|< \delta \Rightarrow |f(x,y)|< \varepsilon$.
$\Leftrightarrow |\sqrt{(y-x+1)^2}|<\delta \Rightarrow |y^2-(x-1)^2|<\varepsilon \Leftrightarrow y-x+1 < \delta \Rightarrow |y^2-(x-1)^2|< \varepsilon$ which work if I take $\delta=\sqrt \varepsilon$. Hmm no it doesn't work. I must sleep now, will think about it.
If you have any help to provide, it's always a pleasure for me. Don't hesitate.
2. Conventions
Let $|y-(x-1)|<\delta$
Now $|y^2-(x-1)^2|=|y-(x-1)||y+(x-1)|<$ $\delta|y-(x-1)+2(x-1)|\leq\delta(|y-(x-1)|+|2(x-1)|)<\delta(\delta+2|x-1|)$
So given an $\epsilon$, $\delta$ must satisfy $\delta^2+2\delta|x-1|<\epsilon$. Since it is dependent on x, this can never be achieved.
Visually speaking, the higher x goes, the steeper the graph becomes when crossing into A. As x goes to infinity, this steepness tends asymptotically to a $90^\circ$ fold in the graph. Therefore given a height variation $\pm\epsilon$, there can never exist a circle in the xy-plane of any radius $\delta$ that can trap an area of the graph with no height variations less than $\pm\epsilon$. You can always proceed Northeast along the line $y=x-1$ and find a spot a little bit steeper.
I know this doesn't count as a formal proof. What is the conventional way to disprove continuity via epsilon/delta?
3. Originally Posted by arbolis
Let $A=\{ (x,y) \in \mathbb{R}^2 :y \leq x-1 \}$ and let $f(x,y)$ be defined in all $\mathbb{R}^2$ by $f(x,y)= \begin{cases} y^2-(x-1)^2 \hspace{2cm} \text{if (x,y)} \in A \\ 0 \hspace{3 cm} \text{if (x,y) is not in A} \end{cases}$.
1)Analyze the continuity of $f$ in all $\mathbb{R}^2$.
I'll put the other questions if I need help, but I'd like a confirmation of my result for part 1).
My attempt :
First I tried to visualize the function and it was quite hard. I couldn't even do that. I realize that it takes the element $y=x-1$ (a line) from the domain and send it to the image as the line $y=0$ (or curve?). I also realize that it sends all the elements under the curve $y=x-1$ into another area, but I didn't visualize it. Anyway, back to the question, I think the limit exists by intuition.
Here's my work to support this : $f$ is continuous in all $\mathbb{R}^2$ except maybe in $y=x-1$.
Right. But you could have been a bit more explicit here. For example you could have argued like this:
A is the half-plane below (and including) the line $y=x-1$. f happens to be continuous at any inner point of either A or $\mathbb{R}^2\backslash A$: because f happens to be identical to a continuous function in an entire neighborhood of such a point.
Thus, the only problematic points are those on the line $y=x-1$.
But note: by arguing like this you already assume that f restricted to A happens to be continuous, so that there is no reason to even try to prove it only for points on the line $y=x-1$ the way you do below:
I'll try to prove it is :
I want to prove that $\lim _{y \to x-1} f(x,y)=0$.
$\forall \varepsilon >0$, $\exists \delta (\varepsilon)$ such that if $|(x,y)-(x,x-1)|< \delta \Rightarrow |f(x,y)|< \varepsilon$.
$\Leftrightarrow |\sqrt{(y-x+1)^2}|<\delta \Rightarrow |y^2-(x-1)^2|<\varepsilon \Leftrightarrow y-x+1 < \delta \Rightarrow |y^2-(x-1)^2|< \varepsilon$ which work if I take $\delta=\sqrt \varepsilon$. Hmm no it doesn't work. I must sleep now, will think about it.
If you have any help to provide, it's always a pleasure for me. Don't hesitate.
Maybe you find it difficult to prove that $f(x,y)$ goes to $f(x_0,y_0)$ as $(x,y)\in A$ tends to $(x_0,y_0)\in A$. Initially I wouldn't have thought of even trying to prove this: one usually assumes that a polynomial function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ of the coordinates of a point of $\mathbb{R}^n$ is continuous without further addo. This is because the projection of a point of $\mathbb{R}^n$ onto its k-th coordinate is continuous, and that sums and products of continous functions are continuous.
Thus there is no reason to prove continuity of a polynomial in the coordinates for each and every such polynomial using a pure delta-epsilon argument: it would be too much work (depending on the complexity of the polynomial) and it wouldn't be particularly enlightening either.
4. (Quoting does not work, sorry).
To Media_Man, ok so if I understand well $f$ is not continuous at $y=x-1$ hence not continuous in all $\mathbb{R}^2$.
To Failure : ok, so if I understand well $f$ is continuous in all $\mathbb{R}^2$.
Originally Posted by Media_Man
I know this doesn't count as a formal proof. What is the conventional way to disprove continuity via epsilon/delta?
with a 2 variables function, we generally approach the limit via different curves. If the limits are different, the real limit does not exist.
5. Allow a point $(x,y)$ to approach the line $y=x-1$ via the following curve: $y=\frac{x^2-2}{x+1}$ (whose slant asymptote is $y=x-1$). This curve is also completely in A. However, $\lim_{x\to\infty}f(x,y)=\lim_{x\to\infty}y^2-(x-1)^2=\lim_{x\to\infty}\left(\frac{x^2-2}{x+1}\right)^2-(x-1)^2=-4\neq 0$
This thread seems confusing to me, as visualizing this function, it seems perfectly continuous by the traditional (intuitive) definition that $\lim_{x\to a}f(x)=f(a)$ for all a.
6. Originally Posted by Media_Man
Allow a point $(x,y)$ to approach the line $y=x-1$ via the following curve: $y=\frac{x^2-2}{x+1}$ (whose slant asymptote is $y=x-1$). This curve is also completely in A. However, $\lim_{x\to\infty}f(x,y)=\lim_{x\to\infty}y^2-(x-1)^2=\lim_{x\to\infty}\left(\frac{x^2-2}{x+1}\right)^2-(x-1)^2=-4\neq 0$
This thread seems confusing to me, as visualizing this function, it seems perfectly continuous by the traditional (intuitive) definition that $\lim_{x\to a}f(x)=f(a)$ for all a.
Let it be noted that the definition of f being continuous at point a via the condition that $\lim_{x\to a}f(x)=f(a)$ holds is not at all merely "intuitive": it is perfectly valid.
The problem, as I see it, is that your method of "disproving continuity" of f does not work. To disprove continuity of f you must show that if $(x,y)$ approaches a specific point $(x_0,y_0)$ in a certain way, the values $f(x,y)$ do not approach $f(x_0,y_0)$. Your letting x tend to infinity, while requiring that $y=\frac{x^2-2}{x+1}$, is definitely not a case of approaching a specific point $(x_0,y_0)$ of the domain of f as required to refute continuity at $(x_0,y_0)$.
7. To Failure : ok, so if I understand well $f$ is continuous in all $\mathbb{R}^2$.
Well, yes: at least that's what I argued. The question is whether you agree with that argument or not.
Of course, you could also use an epsilon-delta style argument to show that $(x,y)\mapsto y^2-(x-1)^2$ is continuous at an arbitrary point $(x_0,y_0)$. However, this requires a bit of grunt work. Just consider the following: assume $\varepsilon>0$ is given, the question is whether it is possible to find a $\delta >0$ such that $\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ implies that
$|f(x,y)-f(x_0,y_0)|=|(y^2-(x-1)^2)-(y_0^2-(x_0-1)^2)|<\varepsilon$
Using the fact that $a^2-a_0^2=(a-a_0)(a+a_0)$ and the triangle inequality it is possible to specify such a $\delta >0$, but the whole thing is just way too complicated to write up (for an impatent fellow like me at least) since, as I wrote in my first post in this thread, all functions $f:\mathbb{R}^n\rightarrow \mathbb{R}$ whose function term is defined as a polynomial in the coordinates of its argument is continuous because: all projections $(x_1,\ldots,x_n)\mapsto x_k$ are continous and the product and the sum of continuous functions are continuous. (I forgot to mention the trivial case: the constant function must also be known to be continuous.) Every "polynomial f in the coordinates of its argument" can be gotten by adding and multiplying suitably chosen functions that are already known to be continuous (i.e. projections onto coordinate subspaces and the constant functions).
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Question
# In exchange for a \$400 million fixed commitment line of credit, your firm has agreed to...
In exchange for a \$400 million fixed commitment line of credit, your firm has agreed to do the following:
1. Pay 1.97 percent per quarter on any funds actually borrowed.
2. Maintain a 1 percent compensating balance on any funds actually borrowed.
3. Pay an up-front commitment fee of 0.23 percent of the amount of the line.
Based on this information, answer the following:
a. Ignoring the commitment fee, what is the effective annual interest rate on this line of credit? (Do not round intermediate calculations and enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.)
Effective annual rate %
b. Suppose your firm immediately uses \$227 million of the line and pays it off in one year. What is the effective annual interest rate on this \$227 million loan? (Do not round intermediate calculations and enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.)
Effective annual rate %
a). Effective Annual Interest Rate = [(1 + Interest per period)^(Number of compounding periods in a year)] -1
= [1.0197]4 - 1 = 1.0812 - 1 = 0.0812, or 8.12%
b). 5% Compensating Balance = \$227 x 0.05 = \$11.35 million
Upfront Commitment Fee = 0.23% x 227 = \$522,100
Amount Availed = Loan Amount - Compensating Balance - Upfront Commitment Fee
= \$227,000,000 - \$11,350,000 - \$522,100 = \$215,127,900
Interest Cost = \$227,000,000 * 8.12% = \$18,423,154.77
EAR = \$18,423,154.77 / \$215,127,900 = 0.0856, or 8.56%
#### Earn Coins
Coins can be redeemed for fabulous gifts.
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# Statistical Data Modeling¶
Some or most of you have probably taken some undergraduate- or graduate-level statistics courses. Unfortunately, the curricula for most introductory statisics courses are mostly focused on conducting statistical hypothesis tests as the primary means for interest: t-tests, chi-squared tests, analysis of variance, etc. Such tests seek to esimate whether groups or effects are "statistically significant", a concept that is poorly understood, and hence often misused, by most practioners. Even when interpreted correctly, statistical significance is a questionable goal for statistical inference, as it is of limited utility.
A far more powerful approach to statistical analysis involves building flexible models with the overarching aim of estimating quantities of interest. This section of the tutorial illustrates how to use Python to build statistical models of low to moderate difficulty from scratch, and use them to extract estimates and associated measures of uncertainty.
In [1]:
import numpy as np
import pandas as pd
# Set some Pandas options
pd.set_option('display.notebook_repr_html', False)
pd.set_option('display.max_columns', 20)
pd.set_option('display.max_rows', 25)
# Estimation¶
An recurring statistical problem is finding estimates of the relevant parameters that correspond to the distribution that best represents our data.
In parametric inference, we specify a priori a suitable distribution, then choose the parameters that best fit the data.
• e.g. $\mu$ and $\sigma^2$ in the case of the normal distribution
In [2]:
x = array([ 1.00201077, 1.58251956, 0.94515919, 6.48778002, 1.47764604,
5.18847071, 4.21988095, 2.85971522, 3.40044437, 3.74907745,
1.18065796, 3.74748775, 3.27328568, 3.19374927, 8.0726155 ,
0.90326139, 2.34460034, 2.14199217, 3.27446744, 3.58872357,
1.20611533, 2.16594393, 5.56610242, 4.66479977, 2.3573932 ])
_ = hist(x, bins=8)
### Fitting data to probability distributions¶
We start with the problem of finding values for the parameters that provide the best fit between the model and the data, called point estimates. First, we need to define what we mean by ‘best fit’. There are two commonly used criteria:
• Method of moments chooses the parameters so that the sample moments (typically the sample mean and variance) match the theoretical moments of our chosen distribution.
• Maximum likelihood chooses the parameters to maximize the likelihood, which measures how likely it is to observe our given sample.
### Discrete Random Variables¶
$$X = \{0,1\}$$$$Y = \{\ldots,-2,-1,0,1,2,\ldots\}$$
Probability Mass Function:
For discrete $X$,
$$Pr(X=x) = f(x|\theta)$$
e.g. Poisson distribution
The Poisson distribution models unbounded counts:
$$Pr(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$$
• $X=\{0,1,2,\ldots\}$
• $\lambda > 0$
$$E(X) = \text{Var}(X) = \lambda$$
### Continuous Random Variables¶
$$X \in [0,1]$$$$Y \in (-\infty, \infty)$$
Probability Density Function:
For continuous $X$,
$$Pr(x \le X \le x + dx) = f(x|\theta)dx \, \text{ as } \, dx \rightarrow 0$$
e.g. normal distribution
$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]$$
• $X \in \mathbf{R}$
• $\mu \in \mathbf{R}$
• $\sigma>0$
\begin{align}E(X) &= \mu \cr \text{Var}(X) &= \sigma^2 \end{align}
### Example: Nashville Precipitation¶
The dataset nashville_precip.txt contains NOAA precipitation data for Nashville measured since 1871. The gamma distribution is often a good fit to aggregated rainfall data, and will be our candidate distribution in this case.
In [3]:
precip = pd.read_table("data/nashville_precip.txt", index_col=0, na_values='NA', delim_whitespace=True)
Out[3]:
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Year
1871 2.76 4.58 5.01 4.13 3.30 2.98 1.58 2.36 0.95 1.31 2.13 1.65
1872 2.32 2.11 3.14 5.91 3.09 5.17 6.10 1.65 4.50 1.58 2.25 2.38
1873 2.96 7.14 4.11 3.59 6.31 4.20 4.63 2.36 1.81 4.28 4.36 5.94
1874 5.22 9.23 5.36 11.84 1.49 2.87 2.65 3.52 3.12 2.63 6.12 4.19
1875 6.15 3.06 8.14 4.22 1.73 5.63 8.12 1.60 3.79 1.25 5.46 4.30
In [4]:
_ = precip.hist(sharex=True, sharey=True, grid=False)
tight_layout()
The first step is recognixing what sort of distribution to fit our data to. A couple of observations:
1. The data are skewed, with a longer tail to the right than to the left
2. The data are positive-valued, since they are measuring rainfall
3. The data are continuous
There are a few possible choices, but one suitable alternative is the gamma distribution:
$$x \sim \text{Gamma}(\alpha, \beta) = \frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}$$
The method of moments simply assigns the empirical mean and variance to their theoretical counterparts, so that we can solve for the parameters.
So, for the gamma distribution, the mean and variance are:
$$\hat{\mu} = \bar{X} = \alpha \beta$$ $$\hat{\sigma}^2 = S^2 = \alpha \beta^2$$
So, if we solve for these parameters, we can use a gamma distribution to describe our data:
$$\alpha = \frac{\bar{X}^2}{S^2}, \, \beta = \frac{S^2}{\bar{X}}$$
Let's deal with the missing value in the October data. Given what we are trying to do, it is most sensible to fill in the missing value with the average of the available values.
In [5]:
precip.fillna(value={'Oct': precip.Oct.mean()}, inplace=True)
Out[5]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 141 entries, 1871 to 2011
Data columns (total 12 columns):
Jan 141 non-null values
Feb 141 non-null values
Mar 141 non-null values
Apr 141 non-null values
May 141 non-null values
Jun 141 non-null values
Jul 141 non-null values
Aug 141 non-null values
Sep 141 non-null values
Oct 141 non-null values
Nov 141 non-null values
Dec 141 non-null values
dtypes: float64(12)
Now, let's calculate the sample moments of interest, the means and variances by month:
In [6]:
precip_mean = precip.mean()
precip_mean
Out[6]:
Jan 4.523688
Feb 4.097801
Mar 4.977589
Apr 4.204468
May 4.325674
Jun 3.873475
Jul 3.895461
Aug 3.367305
Sep 3.377660
Oct 2.610500
Nov 3.685887
Dec 4.176241
dtype: float64
In [7]:
precip_var = precip.var()
precip_var
Out[7]:
Jan 6.928862
Feb 5.516660
Mar 5.365444
Apr 4.117096
May 5.306409
Jun 5.033206
Jul 3.777012
Aug 3.779876
Sep 4.940099
Oct 2.741659
Nov 3.679274
Dec 5.418022
dtype: float64
We then use these moments to estimate $\alpha$ and $\beta$ for each month:
In [8]:
alpha_mom = precip_mean ** 2 / precip_var
beta_mom = precip_var / precip_mean
In [9]:
alpha_mom, beta_mom
Out[9]:
(Jan 2.953407
Feb 3.043866
Mar 4.617770
Apr 4.293694
May 3.526199
Jun 2.980965
Jul 4.017624
Aug 2.999766
Sep 2.309383
Oct 2.485616
Nov 3.692511
Dec 3.219070
dtype: float64,
Jan 1.531684
Feb 1.346249
Mar 1.077920
Apr 0.979219
May 1.226724
Jun 1.299403
Jul 0.969593
Aug 1.122522
Sep 1.462581
Oct 1.050243
Nov 0.998206
Dec 1.297344
dtype: float64)
We can use the gamma.pdf function in scipy.stats.distributions to plot the ditribtuions implied by the calculated alphas and betas. For example, here is January:
In [10]:
from scipy.stats.distributions import gamma
hist(precip.Jan, normed=True, bins=20)
plot(linspace(0, 10), gamma.pdf(linspace(0, 10), alpha_mom[0], beta_mom[0]))
Out[10]:
[<matplotlib.lines.Line2D at 0x113540450>]
Looping over all months, we can create a grid of plots for the distribution of rainfall, using the gamma distribution:
In [11]:
axs = precip.hist(normed=True, figsize=(12, 8), sharex=True, sharey=True, bins=15, grid=False)
for ax in axs.ravel():
# Get month
m = ax.get_title()
# Plot fitted distribution
x = linspace(*ax.get_xlim())
ax.plot(x, gamma.pdf(x, alpha_mom[m], beta_mom[m]))
# Annotate with parameter estimates
label = 'alpha = {0:.2f}\nbeta = {1:.2f}'.format(alpha_mom[m], beta_mom[m])
ax.annotate(label, xy=(10, 0.2))
tight_layout()
# Maximum Likelihood¶
Maximum likelihood (ML) fitting is usually more work than the method of moments, but it is preferred as the resulting estimator is known to have good theoretical properties.
There is a ton of theory regarding ML. We will restrict ourselves to the mechanics here.
Say we have some data $y = y_1,y_2,\ldots,y_n$ that is distributed according to some distribution:
$$Pr(Y_i=y_i | \theta)$$
Here, for example, is a Poisson distribution that describes the distribution of some discrete variables, typically counts:
In [12]:
y = np.random.poisson(5, size=100)
plt.hist(y, bins=12, normed=True)
xlabel('y'); ylabel('Pr(y)')
Out[12]:
<matplotlib.text.Text at 0x11367bf10>
The product $\prod_{i=1}^n Pr(y_i | \theta)$ gives us a measure of how likely it is to observe values $y_1,\ldots,y_n$ given the parameters $\theta$. Maximum likelihood fitting consists of choosing the appropriate function $l= Pr(Y|\theta)$ to maximize for a given set of observations. We call this function the likelihood function, because it is a measure of how likely the observations are if the model is true.
Given these data, how likely is this model?
In the above model, the data were drawn from a Poisson distribution with parameter $\lambda =5$.
$$L(y|\lambda=5) = \frac{e^{-5} 5^y}{y!}$$
So, for any given value of $y$, we can calculate its likelihood:
In [13]:
poisson_like = lambda x, lam: np.exp(-lam) * (lam**x) / (np.arange(x)+1).prod()
lam = 6
value = 10
poisson_like(value, lam)
Out[13]:
0.041303093412337726
In [14]:
np.sum(poisson_like(yi, lam) for yi in y)
Out[14]:
11.968386407200342
In [15]:
lam = 8
np.sum(poisson_like(yi, lam) for yi in y)
Out[15]:
8.5185194171986929
We can plot the likelihood function for any value of the parameter(s):
In [16]:
lambdas = np.linspace(0,15)
x = 5
plt.plot(lambdas, [poisson_like(x, l) for l in lambdas])
xlabel('$\lambda$')
ylabel('L($\lambda$|x={0})'.format(x))
Out[16]:
<matplotlib.text.Text at 0x113c7ebd0>
How is the likelihood function different than the probability distribution function (PDF)? The likelihood is a function of the parameter(s) given the data, whereas the PDF returns the probability of data given a particular parameter value. Here is the PDF of the Poisson for $\lambda=5$.
In [17]:
lam = 5
xvals = arange(15)
plt.bar(xvals, [poisson_like(x, lam) for x in xvals])
xlabel('x')
ylabel('Pr(X|$\lambda$=5)')
Out[17]:
<matplotlib.text.Text at 0x113d83890>
Why are we interested in the likelihood function?
A reasonable estimate of the true, unknown value for the parameter is one which maximizes the likelihood function. So, inference is reduced to an optimization problem.
Going back to the rainfall data, if we are using a gamma distribution we need to maximize:
\begin{align}l(\alpha,\beta) &= \sum_{i=1}^n \log[\beta^{\alpha} x^{\alpha-1} e^{-x/\beta}\Gamma(\alpha)^{-1}] \cr &= n[(\alpha-1)\overline{\log(x)} - \bar{x}\beta + \alpha\log(\beta) - \log\Gamma(\alpha)]\end{align}
(Its usually easier to work in the log scale)
where $n = 2012 − 1871 = 141$ and the bar indicates an average over all i. We choose $\alpha$ and $\beta$ to maximize $l(\alpha,\beta)$.
Notice $l$ is infinite if any $x$ is zero. We do not have any zeros, but we do have an NA value for one of the October data, which we dealt with above.
### Finding the MLE¶
To find the maximum of any function, we typically take the derivative with respect to the variable to be maximized, set it to zero and solve for that variable.
$$\frac{\partial l(\alpha,\beta)}{\partial \beta} = n\left(\frac{\alpha}{\beta} - \bar{x}\right) = 0$$
Which can be solved as $\beta = \alpha/\bar{x}$. However, plugging this into the derivative with respect to $\alpha$ yields:
$$\frac{\partial l(\alpha,\beta)}{\partial \alpha} = \log(\alpha) + \overline{\log(x)} - \log(\bar{x}) - \frac{\Gamma(\alpha)'}{\Gamma(\alpha)} = 0$$
This has no closed form solution. We must use numerical optimization!
Numerical optimization alogarithms take an initial "guess" at the solution, and iteratively improve the guess until it gets "close enough" to the answer.
Here, we will use Newton-Raphson algorithm:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Which is available to us via SciPy:
In [18]:
from scipy.optimize import newton
Here is a graphical example of how Newtone-Raphson converges on a solution, using an arbitrary function:
In [19]:
# some function
func = lambda x: 3./(1 + 400*np.exp(-2*x)) - 1
xvals = np.linspace(0, 6)
plot(xvals, func(xvals))
text(5.3, 2.1, '$f(x)$', fontsize=16)
# zero line
plot([0,6], [0,0], 'k-')
# value at step n
plot([4,4], [0,func(4)], 'k:')
plt.text(4, -.2, '$x_n$', fontsize=16)
# tangent line
tanline = lambda x: -0.858 + 0.626*x
plot(xvals, tanline(xvals), 'r--')
# point at step n+1
xprime = 0.858/0.626
plot([xprime, xprime], [tanline(xprime), func(xprime)], 'k:')
plt.text(xprime+.1, -.2, '$x_{n+1}$', fontsize=16)
Out[19]:
<matplotlib.text.Text at 0x113f4ec10>
To apply the Newton-Raphson algorithm, we need a function that returns a vector containing the first and second derivatives of the function with respect to the variable of interest. In our case, this is:
In [20]:
from scipy.special import psi, polygamma
dlgamma = lambda m, log_mean, mean_log: np.log(m) - psi(m) - log_mean + mean_log
dl2gamma = lambda m, *args: 1./m - polygamma(1, m)
where log_mean and mean_log are $\log{\bar{x}}$ and $\overline{\log(x)}$, respectively. psi and polygamma are complex functions of the Gamma function that result when you take first and second derivatives of that function.
In [21]:
# Calculate statistics
log_mean = precip.mean().apply(log)
mean_log = precip.apply(log).mean()
Time to optimize!
In [22]:
# Alpha MLE for December
alpha_mle = newton(dlgamma, 2, dl2gamma, args=(log_mean[-1], mean_log[-1]))
alpha_mle
Out[22]:
3.5189679152399647
And now plug this back into the solution for beta:
$$\beta = \frac{\alpha}{\bar{X}}$$
In [23]:
beta_mle = alpha_mle/precip.mean()[-1]
beta_mle
Out[23]:
0.84261607548413797
We can compare the fit of the estimates derived from MLE to those from the method of moments:
In [24]:
dec = precip.Dec
dec.hist(normed=True, bins=10, grid=False)
x = linspace(0, dec.max())
plot(x, gamma.pdf(x, alpha_mom[-1], beta_mom[-1]), 'm-')
plot(x, gamma.pdf(x, alpha_mle, beta_mle), 'r--')
Out[24]:
[<matplotlib.lines.Line2D at 0x113ed5f90>]
For some common distributions, SciPy includes methods for fitting via MLE:
In [25]:
from scipy.stats import gamma
gamma.fit(precip.Dec)
Out[25]:
(2.2427517865962434, 0.65494603858755807, 1.5700739281147422)
This fit is not directly comparable to our estimates, however, because SciPy's gamma.fit method fits an odd 3-parameter version of the gamma distribution.
### Example: truncated distribution¶
Suppose that we observe $Y$ truncated below at $a$ (where $a$ is known). If $X$ is the distribution of our observation, then:
$$P(X \le x) = P(Y \le x|Y \gt a) = \frac{P(a \lt Y \le x)}{P(Y \gt a)}$$
(so, $Y$ is the original variable and $X$ is the truncated variable)
Then X has the density:
$$f_X(x) = \frac{f_Y (x)}{1−F_Y (a)} \, \text{for} \, x \gt a$$
Suppose $Y \sim N(\mu, \sigma^2)$ and $x_1,\ldots,x_n$ are independent observations of $X$. We can use maximum likelihood to find $\mu$ and $\sigma$.
First, we can simulate a truncated distribution using a while statement to eliminate samples that are outside the support of the truncated distribution.
In [26]:
x = np.random.normal(size=10000)
a = -1
x_small = x < a
while x_small.sum():
x[x_small] = np.random.normal(size=x_small.sum())
x_small = x < a
_ = hist(x, bins=100)
We can construct a log likelihood for this function using the conditional form:
$$f_X(x) = \frac{f_Y (x)}{1−F_Y (a)} \, \text{for} \, x \gt a$$
In [27]:
from scipy.stats.distributions import norm
trunc_norm = lambda theta, a, x: -(np.log(norm.pdf(x, theta[0], theta[1])) -
np.log(1 - norm.cdf(a, theta[0], theta[1]))).sum()
For this example, we will use another optimization algorithm, the Nelder-Mead simplex algorithm. It has a couple of advantages:
• it does not require derivatives
• it can optimize (minimize) a vector of parameters
SciPy implements this algorithm in its fmin function:
In [28]:
from scipy.optimize import fmin
fmin(trunc_norm, np.array([1,2]), args=(-1, x))
Optimization terminated successfully.
Current function value: 10930.807932
Iterations: 47
Function evaluations: 89
Out[28]:
array([-0.01353635, 0.9928185 ])
In general, simulating data is a terrific way of testing your model before using it with real data.
### Kernel density estimates¶
In some instances, we may not be interested in the parameters of a particular distribution of data, but just a smoothed representation of the data at hand. In this case, we can estimate the disribution non-parametrically (i.e. making no assumptions about the form of the underlying distribution) using kernel density estimation.
In [29]:
# Some random data
y = np.random.random(15) * 10
y
Out[29]:
array([ 4.74511593, 7.90840801, 0.4757153 , 2.20288601, 1.84984043,
7.10173654, 2.44884627, 1.34956892, 6.86027822, 3.71783498,
7.42356175, 4.07499025, 1.42352956, 3.6932789 , 9.22449069])
In [30]:
x = np.linspace(0, 10, 100)
# Smoothing parameter
s = 0.4
# Calculate the kernels
kernels = np.transpose([norm.pdf(x, yi, s) for yi in y])
plot(x, kernels, 'k:')
plot(x, kernels.sum(1))
plot(y, np.zeros(len(y)), 'ro', ms=10)
Out[30]:
[<matplotlib.lines.Line2D at 0x11400ae50>]
SciPy implements a Gaussian KDE that automatically chooses an appropriate bandwidth. Let's create a bi-modal distribution of data that is not easily summarized by a parametric distribution:
In [31]:
# Create a bi-modal distribution with a mixture of Normals.
x1 = np.random.normal(0, 3, 50)
x2 = np.random.normal(4, 1, 50)
# Append by row
x = np.r_[x1, x2]
In [32]:
plt.hist(x, bins=8, normed=True)
Out[32]:
(array([ 0.03623044, 0.04830725, 0.06038407, 0.07246088, 0.03623044,
0.15096017, 0.18719061, 0.01207681]),
array([-6.09280539, -4.4367394 , -2.78067341, -1.12460742, 0.53145857,
2.18752456, 3.84359055, 5.49965654, 7.15572253]),
<a list of 8 Patch objects>)
In [33]:
from scipy.stats import kde
density = kde.gaussian_kde(x)
xgrid = np.linspace(x.min(), x.max(), 100)
plt.hist(x, bins=8, normed=True)
plt.plot(xgrid, density(xgrid), 'r-')
Out[33]:
[<matplotlib.lines.Line2D at 0x114129fd0>]
### Exercise: Cervical dystonia analysis¶
Recall the cervical dystonia database, which is a clinical trial of botulinum toxin type B (BotB) for patients with cervical dystonia from nine U.S. sites. The response variable is measurements on the Toronto Western Spasmodic Torticollis Rating Scale (TWSTRS), measuring severity, pain, and disability of cervical dystonia (high scores mean more impairment). One way to check the efficacy of the treatment is to compare the distribution of TWSTRS for control and treatment patients at the end of the study.
Use the method of moments or MLE to calculate the mean and variance of TWSTRS at week 16 for one of the treatments and the control group. Assume that the distribution of the twstrs variable is normal:
$$f(x \mid \mu, \sigma^2) = \sqrt{\frac{1}{2\pi\sigma^2}} \exp\left\{ -\frac{1}{2} \frac{(x-\mu)^2}{\sigma^2} \right\}$$
In [34]:
cdystonia = pd.read_csv("data/cdystonia.csv")
cdystonia[cdystonia.obs==6].hist(column='twstrs', by=cdystonia.treat, bins=8)
Out[34]:
array([[<matplotlib.axes.AxesSubplot object at 0x1140d01d0>,
<matplotlib.axes.AxesSubplot object at 0x11409e810>],
[<matplotlib.axes.AxesSubplot object at 0x11427dc90>,
<matplotlib.axes.AxesSubplot object at 0x114299f50>]], dtype=object)
In [34]:
## Regression models¶
A general, primary goal of many statistical data analysis tasks is to relate the influence of one variable on another. For example, we may wish to know how different medical interventions influence the incidence or duration of disease, or perhaps a how baseball player's performance varies as a function of age.
In [35]:
x = np.array([2.2, 4.3, 5.1, 5.8, 6.4, 8.0])
y = np.array([0.4, 10.1, 14.0, 10.9, 15.4, 18.5])
plot(x,y,'ro')
Out[35]:
[<matplotlib.lines.Line2D at 0x1143406d0>]
We can build a model to characterize the relationship between $X$ and $Y$, recognizing that additional factors other than $X$ (the ones we have measured or are interested in) may influence the response variable $Y$.
$y_i = f(x_i) + \epsilon_i$
where $f$ is some function, for example a linear function:
$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$
and $\epsilon_i$ accounts for the difference between the observed response $y_i$ and its prediction from the model $\hat{y_i} = \beta_0 + \beta_1 x_i$. This is sometimes referred to as process uncertainty.
We would like to select $\beta_0, \beta_1$ so that the difference between the predictions and the observations is zero, but this is not usually possible. Instead, we choose a reasonable criterion: the smallest sum of the squared differences between $\hat{y}$ and $y$.
$$R^2 = \sum_i (y_i - [\beta_0 + \beta_1 x_i])^2 = \sum_i \epsilon_i^2$$
Squaring serves two purposes: (1) to prevent positive and negative values from cancelling each other out and (2) to strongly penalize large deviations. Whether the latter is a good thing or not depends on the goals of the analysis.
In other words, we will select the parameters that minimize the squared error of the model.
In [36]:
ss = lambda theta, x, y: np.sum((y - theta[0] - theta[1]*x) ** 2)
In [37]:
ss([0,1],x,y)
Out[37]:
333.35000000000002
In [38]:
b0,b1 = fmin(ss, [0,1], args=(x,y))
b0,b1
Optimization terminated successfully.
Current function value: 21.375000
Iterations: 79
Function evaluations: 153
Out[38]:
(-4.3500136038870876, 3.0000002915386412)
In [39]:
plot(x, y, 'ro')
plot([0,10], [b0, b0+b1*10])
Out[39]:
[<matplotlib.lines.Line2D at 0x113f66410>]
In [40]:
plot(x, y, 'ro')
plot([0,10], [b0, b0+b1*10])
for xi, yi in zip(x,y):
plot([xi]*2, [yi, b0+b1*xi], 'k:')
xlim(2, 9); ylim(0, 20)
Out[40]:
(0, 20)
Minimizing the sum of squares is not the only criterion we can use; it is just a very popular (and successful) one. For example, we can try to minimize the sum of absolute differences:
In [41]:
sabs = lambda theta, x, y: np.sum(np.abs(y - theta[0] - theta[1]*x))
b0,b1 = fmin(sabs, [0,1], args=(x,y))
print b0,b1
plot(x, y, 'ro')
plot([0,10], [b0, b0+b1*10])
Optimization terminated successfully.
Current function value: 10.162463
Iterations: 39
Function evaluations: 77
0.00157170444494 2.31231743181
Out[41]:
[<matplotlib.lines.Line2D at 0x11438fad0>]
We are not restricted to a straight-line regression model; we can represent a curved relationship between our variables by introducing polynomial terms. For example, a cubic model:
$y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + \epsilon_i$
In [42]:
ss2 = lambda theta, x, y: np.sum((y - theta[0] - theta[1]*x - theta[2]*(x**2)) ** 2)
b0,b1,b2 = fmin(ss2, [1,1,-1], args=(x,y))
print b0,b1,b2
plot(x, y, 'ro')
xvals = np.linspace(0, 10, 100)
plot(xvals, b0 + b1*xvals + b2*(xvals**2))
Optimization terminated successfully.
Current function value: 14.001110
Iterations: 198
Function evaluations: 372
-11.0748186039 6.0576975948 -0.302681057088
Out[42]:
[<matplotlib.lines.Line2D at 0x114440890>]
Although polynomial model characterizes a nonlinear relationship, it is a linear problem in terms of estimation. That is, the regression model $f(y | x)$ is linear in the parameters.
For some data, it may be reasonable to consider polynomials of order>2. For example, consider the relationship between the number of home runs a baseball player hits and the number of runs batted in (RBI) they accumulate; clearly, the relationship is positive, but we may not expect a linear relationship.
In [43]:
ss3 = lambda theta, x, y: np.sum((y - theta[0] - theta[1]*x - theta[2]*(x**2) - theta[3]*(x**3)) ** 2)
plot(bb.hr, bb.rbi, 'r.')
b0,b1,b2,b3 = fmin(ss3, [0,1,-1,0], args=(bb.hr, bb.rbi))
xvals = arange(40)
plot(xvals, b0 + b1*xvals + b2*(xvals**2) + b3*(xvals**3))
Optimization terminated successfully.
Current function value: 4274.128398
Iterations: 230
Function evaluations: 407
Out[43]:
[<matplotlib.lines.Line2D at 0x114476750>]
Of course, we need not fit least squares models by hand. The statsmodels package implements least squares models that allow for model fitting in a single line:
In [44]:
import statsmodels.api as sm
straight_line.summary()
Out[44]:
Dep. Variable: R-squared: y 0.891 OLS 0.864 Least Squares 32.67 Mon, 24 Jun 2013 0.00463 21:24:18 -12.325 6 28.65 4 28.23 1
coef std err t P>|t| [95.0% Conf. Int.] -4.3500 2.937 -1.481 0.213 -12.505 3.805 3.0000 0.525 5.716 0.005 1.543 4.457
Omnibus: Durbin-Watson: nan 2.387 nan 0.57 0.359 0.752 1.671 17.9
In [45]:
from statsmodels.formula.api import ols as OLS
data = pd.DataFrame(dict(x=x, x2=x**2, y=y))
cubic_fit = OLS('y ~ x + x2', data).fit()
cubic_fit.summary()
Out[45]:
Dep. Variable: R-squared: y 0.929 OLS 0.881 Least Squares 19.50 Mon, 24 Jun 2013 0.0191 21:24:18 -11.056 6 28.11 3 27.49 2
coef std err t P>|t| [95.0% Conf. Int.] -11.0748 6.013 -1.842 0.163 -30.211 8.062 6.0577 2.482 2.441 0.092 -1.840 13.955 -0.3027 0.241 -1.257 0.298 -1.069 0.464
Omnibus: Durbin-Watson: nan 2.711 nan 0.655 -0.809 0.721 2.961 270
### Exercise: Polynomial function¶
Write a function that specified a polynomial of arbitrary degree.
In [45]:
## Model Selection¶
How do we choose among competing models for a given dataset? More parameters are not necessarily better, from the standpoint of model fit. For example, fitting a 9-th order polynomial to the sample data from the above example certainly results in an overfit.
In [46]:
def calc_poly(params, data):
x = np.c_[[data**i for i in range(len(params))]]
return np.dot(params, x)
ssp = lambda theta, x, y: np.sum((y - calc_poly(theta, x)) ** 2)
betas = fmin(ssp, np.zeros(10), args=(x,y), maxiter=1e6)
plot(x, y, 'ro')
xvals = np.linspace(0, max(x), 100)
plot(xvals, calc_poly(betas, xvals))
Optimization terminated successfully.
Current function value: 7.015262
Iterations: 663
Function evaluations: 983
Out[46]:
[<matplotlib.lines.Line2D at 0x1145b5dd0>]
One approach is to use an information-theoretic criterion to select the most appropriate model. For example Akaike's Information Criterion (AIC) balances the fit of the model (in terms of the likelihood) with the number of parameters required to achieve that fit. We can easily calculate AIC as:
$$AIC = n \log(\hat{\sigma}^2) + 2p$$
where $p$ is the number of parameters in the model and $\hat{\sigma}^2 = RSS/(n-p-1)$.
Notice that as the number of parameters increase, the residual sum of squares goes down, but the second term (a penalty) increases.
To apply AIC to model selection, we choose the model that has the lowest AIC value.
In [47]:
n = len(x)
RSS1 = ss(fmin(ss, [0,1], args=(x,y)), x, y)
RSS2 = ss2(fmin(ss2, [1,1,-1], args=(x,y)), x, y)
Optimization terminated successfully.
Current function value: 21.375000
Iterations: 79
Function evaluations: 153
Optimization terminated successfully.
Current function value: 14.001110
Iterations: 198
Function evaluations: 372
15.7816583572 17.6759368019
Hence, we would select the 2-parameter (linear) model.
## Logistic Regression¶
Fitting a line to the relationship between two variables using the least squares approach is sensible when the variable we are trying to predict is continuous, but what about when the data are dichotomous?
• male/female
• pass/fail
• died/survived
Let's consider the problem of predicting survival in the Titanic disaster, based on our available information. For example, lets say that we want to predict survival as a function of the fare paid for the journey.
In [48]:
titanic = pd.read_excel("data/titanic.xls", "titanic")
titanic.name
Out[48]:
0 Allen, Miss. Elisabeth Walton
1 Allison, Master. Hudson Trevor
2 Allison, Miss. Helen Loraine
3 Allison, Mr. Hudson Joshua Creighton
4 Allison, Mrs. Hudson J C (Bessie Waldo Daniels)
5 Anderson, Mr. Harry
6 Andrews, Miss. Kornelia Theodosia
7 Andrews, Mr. Thomas Jr
8 Appleton, Mrs. Edward Dale (Charlotte Lamson)
9 Artagaveytia, Mr. Ramon
...
1298 Wittevrongel, Mr. Camille
1299 Yasbeck, Mr. Antoni
1300 Yasbeck, Mrs. Antoni (Selini Alexander)
1301 Youseff, Mr. Gerious
1302 Yousif, Mr. Wazli
1303 Yousseff, Mr. Gerious
1304 Zabour, Miss. Hileni
1305 Zabour, Miss. Thamine
1306 Zakarian, Mr. Mapriededer
1307 Zakarian, Mr. Ortin
1308 Zimmerman, Mr. Leo
Name: name, Length: 1309, dtype: object
In [49]:
jitter = np.random.normal(scale=0.02, size=len(titanic))
plt.scatter(np.log(titanic.fare), titanic.survived + jitter, alpha=0.3)
yticks([0,1])
ylabel("survived")
xlabel("log(fare)")
Out[49]:
<matplotlib.text.Text at 0x1145361d0>
I have added random jitter on the y-axis to help visualize the density of the points, and have plotted fare on the log scale.
Clearly, fitting a line through this data makes little sense, for several reasons. First, for most values of the predictor variable, the line would predict values that are not zero or one. Second, it would seem odd to choose least squares (or similar) as a criterion for selecting the best line.
In [50]:
x = np.log(titanic.fare[titanic.fare>0])
y = titanic.survived[titanic.fare>0]
betas_titanic = fmin(ss, [1,1], args=(x,y))
Optimization terminated successfully.
Current function value: 277.621917
Iterations: 55
Function evaluations: 103
In [51]:
jitter = np.random.normal(scale=0.02, size=len(titanic))
plt.scatter(np.log(titanic.fare), titanic.survived + jitter, alpha=0.3)
yticks([0,1])
ylabel("survived")
xlabel("log(fare)")
plt.plot([0,7], [betas_titanic[0], betas_titanic[0] + betas_titanic[1]*7.])
Out[51]:
[<matplotlib.lines.Line2D at 0x1148a1ed0>]
If we look at this data, we can see that for most values of fare, there are some individuals that survived and some that did not. However, notice that the cloud of points is denser on the "survived" (y=1) side for larger values of fare than on the "died" (y=0) side.
### Stochastic model¶
Rather than model the binary outcome explicitly, it makes sense instead to model the probability of death or survival in a stochastic model. Probabilities are measured on a continuous [0,1] scale, which may be more amenable for prediction using a regression line. We need to consider a different probability model for this exerciese however; let's consider the Bernoulli distribution as a generative model for our data:
$$f(y|p) = p^{y} (1-p)^{1-y}$$
where $y = \{0,1\}$ and $p \in [0,1]$. So, this model predicts whether $y$ is zero or one as a function of the probability $p$. Notice that when $y=1$, the $1-p$ term disappears, and when $y=0$, the $p$ term disappears.
So, the model we want to fit should look something like this:
$$p_i = \beta_0 + \beta_1 x_i + \epsilon_i$$
However, since $p$ is constrained to be between zero and one, it is easy to see where a linear (or polynomial) model might predict values outside of this range. We can modify this model sligtly by using a link function to transform the probability to have an unbounded range on a new scale. Specifically, we can use a logit transformation as our link function:
$$\text{logit}(p) = \log\left[\frac{p}{1-p}\right] = x$$
Here's a plot of $p/(1-p)$
In [52]:
logit = lambda p: np.log(p/(1.-p))
unit_interval = np.linspace(0,1)
plt.plot(unit_interval/(1-unit_interval), unit_interval)
Out[52]:
[<matplotlib.lines.Line2D at 0x114849d50>]
And here's the logit function:
In [53]:
plt.plot(logit(unit_interval), unit_interval)
Out[53]:
[<matplotlib.lines.Line2D at 0x1148baf90>]
The inverse of the logit transformation is:
$$p = \frac{1}{1 + \exp(-x)}$$
So, now our model is:
$$\text{logit}(p_i) = \beta_0 + \beta_1 x_i + \epsilon_i$$
We can fit this model using maximum likelihood. Our likelihood, again based on the Bernoulli model is:
$$L(y|p) = \prod_{i=1}^n p_i^{y_i} (1-p_i)^{1-y_i}$$
which, on the log scale is:
$$l(y|p) = \sum_{i=1}^n y_i \log(p_i) + (1-y_i)\log(1-p_i)$$
We can easily implement this in Python, keeping in mind that fmin minimizes, rather than maximizes functions:
In [54]:
invlogit = lambda x: 1. / (1 + np.exp(-x))
def logistic_like(theta, x, y):
p = invlogit(theta[0] + theta[1] * x)
# Return negative of log-likelihood
return -np.sum(y * np.log(p) + (1-y) * np.log(1 - p))
Remove null values from variables
In [70]:
x, y = titanic[titanic.fare.notnull()][['fare', 'survived']].values.T
... and fit the model.
In [71]:
b0,b1 = fmin(logistic_like, [0.5,0], args=(x,y))
b0, b1
Optimization terminated successfully.
Current function value: 827.015955
Iterations: 47
Function evaluations: 93
Out[71]:
(-0.88238984528338194, 0.012452067664164127)
In [72]:
jitter = np.random.normal(scale=0.01, size=len(x))
plot(x, y+jitter, 'r.', alpha=0.3)
yticks([0,.25,.5,.75,1])
xvals = np.linspace(0, 600)
plot(xvals, invlogit(b0+b1*xvals))
Out[72]:
[<matplotlib.lines.Line2D at 0x114b65a10>]
As with our least squares model, we can easily fit logistic regression models in statsmodels, in this case using the GLM (generalized linear model) class with a binomial error distribution specified.
In [74]:
logistic = sm.GLM(y, sm.add_constant(x), family=sm.families.Binomial()).fit()
logistic.summary()
Out[74]:
Dep. Variable: No. Observations: y 1308 GLM 1306 Binomial 1 logit 1.0 IRLS -827.02 Mon, 24 Jun 2013 1654.0 21:38:17 1.33e+03 6
coef std err t P>|t| [95.0% Conf. Int.] -0.8824 0.076 -11.684 0.000 -1.030 -0.734 0.0125 0.002 7.762 0.000 0.009 0.016
### Exercise: multivariate logistic regression¶
Which other variables might be relevant for predicting the probability of surviving the Titanic? Generalize the model likelihood to include 2 or 3 other covariates from the dataset.
In [74]:
## Bootstrapping¶
Parametric inference can be non-robust:
• inaccurate if parametric assumptions are violated
• if we rely on asymptotic results, we may not achieve an acceptable level of accuracy
Parmetric inference can be difficult:
• derivation of sampling distribution may not be possible
An alternative is to estimate the sampling distribution of a statistic empirically without making assumptions about the form of the population.
We have seen this already with the kernel density estimate.
### Non-parametric Bootstrap¶
The bootstrap is a resampling method discovered by Brad Efron that allows one to approximate the true sampling distribution of a dataset, and thereby obtain estimates of the mean and variance of the distribution.
Bootstrap sample:
$$S_1^* = \{x_{11}^*, x_{12}^*, \ldots, x_{1n}^*\}$$
$S_i^*$ is a sample of size $n$, with replacement.
In Python, we have already seen the NumPy function permutation that can be used in conjunction with Pandas' take method to generate a random sample of some data without replacement:
In [75]:
np.random.permutation(titanic.name)[:5]
Out[75]:
array([u'Honkanen, Miss. Eliina',
u'Andersen-Jensen, Miss. Carla Christine Nielsine',
u'Geiger, Miss. Amalie', u'Becker, Master. Richard F',
u'Johnson, Mr. Malkolm Joackim'], dtype=object)
Similarly, we can use the random.randint method to generate a sample with replacement, which we can use when bootstrapping.
In [76]:
random_ind = np.random.randint(0, len(titanic), 5)
titanic.name[random_ind]
Out[76]:
659 Baclini, Miss. Marie Catherine
1029 Moran, Mr. Daniel J
1129 Petterson, Mr. Johan Emil
556 Sharp, Mr. Percival James R
1045 Myhrman, Mr. Pehr Fabian Oliver Malkolm
Name: name, dtype: object
We regard S as an "estimate" of population P
population : sample :: sample : bootstrap sample
The idea is to generate replicate bootstrap samples:
$$S^* = \{S_1^*, S_2^*, \ldots, S_R^*\}$$
Compute statistic $t$ (estimate) for each bootstrap sample:
$$T_i^* = t(S^*)$$
In [77]:
n = 10
R = 1000
# Original sample (n=10)
x = np.random.normal(size=n)
# 1000 bootstrap samples of size 10
s = [x[np.random.randint(0,n,n)].mean() for i in range(R)]
_ = hist(s, bins=30)
### Bootstrap Estimates¶
From our bootstrapped samples, we can extract estimates of the expectation and its variance:
$$\bar{T}^* = \hat{E}(T^*) = \frac{\sum_i T_i^*}{R}$$$$\hat{\text{Var}}(T^*) = \frac{\sum_i (T_i^* - \bar{T}^*)^2}{R-1}$$
In [78]:
boot_mean = np.sum(s)/R
boot_mean
Out[78]:
-0.15933323217972145
In [79]:
boot_var = ((np.array(s) - boot_mean) ** 2).sum() / (R-1)
boot_var
Out[79]:
0.044061864233632578
Since we have estimated the expectation of the bootstrapped statistics, we can estimate the bias of T:
$$\hat{B}^* = \bar{T}^* - T$$
In [80]:
boot_mean - np.mean(x)
Out[80]:
-0.00067633809837364112
### Bootstrap error¶
There are two sources of error in bootstrap estimates:
1. Sampling error from the selection of $S$.
2. Bootstrap error from failing to enumerate all possible bootstrap samples.
For the sake of accuracy, it is prudent to choose at least R=1000
### Bootstrap Percentile Intervals¶
An attractive feature of bootstrap statistics is the ease with which you can obtain an estimate of uncertainty for a given statistic. We simply use the empirical quantiles of the bootstrapped statistics to obtain percentiles corresponding to a confidence interval of interest.
This employs the ordered bootstrap replicates:
$$T_{(1)}^*, T_{(2)}^*, \ldots, T_{(R)}^*$$
Simply extract the $100(\alpha/2)$ and $100(1-\alpha/2)$ percentiles:
$$T_{[(R+1)\alpha/2]}^* \lt \theta \lt T_{[(R+1)(1-\alpha/2)]}^*$$
In [81]:
s_sorted = np.sort(s)
s_sorted[:10]
Out[81]:
array([-0.84783322, -0.729463 , -0.70891663, -0.70277966, -0.70118232,
-0.69761647, -0.68948506, -0.68308688, -0.67812924, -0.66995858])
In [82]:
s_sorted[-10:]
Out[82]:
array([ 0.28296582, 0.29250432, 0.30557177, 0.31954429, 0.32994401,
0.33001289, 0.33514567, 0.34787314, 0.41920685, 0.48525756])
In [83]:
alpha = 0.05
s_sorted[[(R+1)*alpha/2, (R+1)*(1-alpha/2)]]
Out[83]:
array([-0.55827684, 0.22701275])
### Exercise: Cervical dystonia bootstrap estimates¶
Use bootstrapping to estimate the mean of one of the treatment groups, and calculate percentile intervals for the mean.
In [83]:
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Many pregnant women suffer from vitamin deficiency, but this is frequently not due to vitamin deficiency in their diets; most often it is because they have higher requirements for vitamins than do the rest of the population.
The best criticism of the reasoning in the statement above is that it
(A) fails to specify the percentage of pregnant women who suffer from vitamin deficiency.
(B) gives insufficient information about why pregnant women have higher vitamin requirements than do other groups.
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency."
(D) provides insufficient information about the incidence of vitamin deficiency in other groups with high vitamin requirements
(E) uses "higher requirements" in an ambiguous manner.
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09 Feb 2008, 09:44
C by POE
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency."
Good question (+1), I'm not sure but I would pick C on exam day. What is OA?
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09 Feb 2008, 10:06
IMO: E
[quote"Tom Badgerlock"]What the...? I do not understand the answer for this one!
*******
Many pregnant women suffer from vitamin deficiency, but this is frequently not due to vitamin deficiency in their diets; most often it is because they have higher requirements for vitamins than do the rest of the population.
The best criticism of the reasoning in the statement above is that it
(A) fails to specify the percentage of pregnant women who suffer from vitamin deficiency. [color=#0000FF]Irrelevant, as it implictly assumes the conclusion to be accurate
(B) gives insufficient information about why pregnant women have higher vitamin requirements than do other groups.irrelevant, for the same reason as A
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency."false, as the pronoun "their" clearly refers to "Many pregnant women"
(D) provides insufficient information about the incidence of vitamin deficiency in other groups with high vitamin requirements. irrelevant, as knowing anything about the other group doesn't really tell us anything about our group
(E) uses "higher requirements" in an ambiguous manner. the argument is that -the deficiency is not because of deficiency but because of higher requirements. we don't know anything about the 'higher requirements' so we can't jugde the efficacy of the claim.
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09 Feb 2008, 13:04
What the...? I do not understand the answer for this one!
*******
Many pregnant women suffer from vitamin deficiency, but this is frequently not due to vitamin deficiency in their diets; most often it is because they have higher requirements for vitamins than do the rest of the population.
The best criticism of the reasoning in the statement above is that it
(A) fails to specify the percentage of pregnant women who suffer from vitamin deficiency. Irrelevant
(B) gives insufficient information about why pregnant women have higher vitamin requirements than do other groups. Irrelevant
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency." Correct. This is ambiguous. If they have higher requirements for vitamins, their diet would be lacking in vitamins otherwise.. why would they be deficient? The context for 'vitamin deficiency' as applied to pregnant women is different from the one that's applicable to their diets (which is the standard for the rest of the population)
(D) provides insufficient information about the incidence of vitamin deficiency in other groups with high vitamin requirements Irrelevant
(E) uses "higher requirements" in an ambiguous manner.Incorrect. Higher requirements for vitamin intake is is compared to the rest of the population, and should suffice
The answer should be (C). Whats OA?
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09 Feb 2008, 13:36
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jay02 wrote:
IMO: E
...
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency."false, as the pronoun "their" clearly refers to "Many pregnant women"
...
Many pregnant women suffer from vitamin deficiency - vitamin deficiency: the organism requires more vitamins than it has.
vitamin deficiency in their diets - what does "vitamin deficiency" mean here? or refer here? Clearly, there is other "vitamin deficiency" that applies for diets.
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09 Feb 2008, 19:45
Thanks Walker.
My initial choice was E, but C makes sense.
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09 Feb 2008, 21:06
The OA is C. Thanks for your explanations and contributions, everyone! I'm learning a lot
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10 Feb 2008, 19:57
walker wrote:
jay02 wrote:
IMO: E
...
(C) fails to employ the same reference group for both uses of the term "vitamin deficiency."false, as the pronoun "their" clearly refers to "Many pregnant women"
...
Many pregnant women suffer from vitamin deficiency - vitamin deficiency: the organism requires more vitamins than it has.
vitamin deficiency in their diets - what does "vitamin deficiency" mean here? or refer here? Clearly, there is other "vitamin deficiency" that applies for diets.
Thanks Walker...this one completely stumped me
+1
Re: CR pregnant women [#permalink] 10 Feb 2008, 19:57
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# Find x in infinite sequence [duplicate]
This question already has an answer here:
We have: $$x^{x^{x^{ x^{x ^{x ^{\dots}}}}}} = 2.$$
I tried a reasoning by recursion:
For $$n=1$$: \begin{align} x^x &= 2 \\ \implies x\ln x &= \ln 2 \end{align}
For $$n=2$$: \begin{align} x^{x^x} &= 2 \\ \implies x\ln x^x &= \ln 2 \\ \implies x^2\ln x &= \ln 2 \\ \end{align}
For an arbitrary $$n$$ we can solve: $$x^n \ln x = \ln 2$$
But for $$n \to \infty$$, I can't seem to find something. Does anyone have an idea on how to tackle this problem?
## marked as duplicate by Martin R, Community♦Apr 18 at 7:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• – Martin R Apr 18 at 7:38
## 1 Answer
Hint: If the solution exists then
$$x^{x^{x^{\cdots}}}=x^{(x^{x^{\cdots}})}=x^{(2)}=x^2=2.$$
The stuff in the power is equal to $$2$$ hence we can replace it by $$2$$.
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# 10 topics : how many days until sept 29 ?
Rate this post
Outline:
I. Introduction
– Explanation of the importance of knowing how many days until Sept 29
– Brief overview of the article
II. How to Calculate the Number of Days Until Sept 29
– Explanation of the formula for calculating the number of days
– Step-by-step guide on how to calculate the number of days until Sept 29
– Examples of how to use the formula
III. Why Knowing How Many Days Until Sept 29 is Important
– Explanation of the significance of Sept 29
– Discussion of events or occasions that are happening on Sept 29
– Importance of planning ahead for these events
IV. Tools for Calculating the Number of Days Until Sept 29
– List of online tools that can be used to calculate the number of days
– Comparison of the features and benefits of each tool
– Step-by-step guide on how to use each tool
– Common questions about calculating the number of days until Sept 29
VI. Conclusion
– Recap of the importance of knowing how many days until Sept 29
– Final thoughts and recommendations
Full Content:
I. Introduction
Knowing how many days until Sept 29 is important for a variety of reasons. Whether you are planning an event, scheduling a vacation, or simply counting down the days until a special occasion, having an accurate count of the number of days can help you stay organized and prepared. In this article, we will discuss how to calculate the number of days until Sept 29, why it is important to know this information, and tools that can be used to make the calculation easier.
II. How to Calculate the Number of Days Until Sept 29
Calculating the number of days until Sept 29 is a simple process that can be done using a basic formula. The formula is as follows:
(Number of days in September) – (Current day of the month) + 29
For example, if today is September 1, the formula would be:
(30) – (1) + 29 = 58
This means that there are 58 days until Sept 29.
To calculate the number of days until Sept 29, follow these steps:
1. Determine the number of days in September. September has 30 days, so this number will remain constant.
2. Determine the current day of the month. This will change depending on when you are making the calculation.
3. Subtract the current day of the month from the number of days in September.
4. Add 29 to the result.
5. The final number is the number of days until Sept 29.
Examples:
– If today is September 10, the calculation would be: (30) – (10) + 29 = 49. There are 49 days until Sept 29.
– If today is September 20, the calculation would be: (30) – (20) + 29 = 39. There are 39 days until Sept 29.
– If today is September 28, the calculation would be: (30) – (28) + 29 = 31. There are 31 days until Sept 29.
III. Why Knowing How Many Days Until Sept 29 is Important
Sept 29 is a significant date for many people, and knowing how many days until this date can help with planning and preparation. Some of the events or occasions that are happening on Sept 29 include:
– National Coffee Day: This day is celebrated annually on Sept 29 and is a great opportunity to enjoy a cup of coffee with friends and family.
– Michaelmas: This is a Christian feast day that is celebrated on Sept 29 in honor of the archangel Michael.
– World Heart Day: This day is observed annually on Sept 29 and is a global initiative to raise awareness about heart health.
In addition to these events, Sept 29 may also be important for personal reasons, such as a birthday, anniversary, or other special occasion. By knowing how many days until Sept 29, you can plan ahead and make sure that you are prepared for these events.
IV. Tools for Calculating the Number of Days Until Sept 29
There are several online tools that can be used to calculate the number of days until Sept 29. Some of the most popular tools include:
1. Time and Date Calculator: This tool allows you to calculate the number of days between two dates, including Sept 29. Simply enter the current date and Sept 29, and the tool will do the rest.
2. Days Between Dates Calculator: This tool is similar to the Time and Date Calculator, but it also allows you to calculate the number of days between two specific dates. This can be useful if you need to know the number of days between two events that are not on Sept 29.
3. Countdown Timer: This tool allows you to create a countdown timer for Sept 29. You can customize the timer with your own text and choose from a variety of designs.
Q: What is the significance of Sept 29?
A: Sept 29 is significant for a variety of reasons, including National Coffee Day, Michaelmas, and World Heart Day. It may also be important for personal reasons, such as a birthday or anniversary.
Q: How do I calculate the number of days until Sept 29?
A: To calculate the number of days until Sept 29, use the formula: (Number of days in September) – (Current day of the month) + 29.
Q: Are there any online tools that can help me calculate the number of days until Sept 29?
A: Yes, there are several online tools that can be used to calculate the number of days until Sept 29, including the Time and Date Calculator, Days Between Dates Calculator, and Countdown Timer.
VI. Conclusion
Knowing how many days until Sept 29 is important for a variety of reasons, including planning events, scheduling vacations, and counting down to special occasions. By using the formula or online tools, you can easily calculate the number of days until Sept 29 and stay organized and prepared. Whether you are celebrating National Coffee Day, Michaelmas, World Heart Day, or a personal milestone, knowing how many days until Sept 29 can help you make the most of these special occasions.
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# Write without absolute value notation: abs(6-5i)
Write without absolute value notation: $|6-5i|$
You can still ask an expert for help
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
• Math expert for every subject
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Step 1
Absolute value of a complex number:
The absolute value of a complex number z=x+iy is given by $|z|=|x+iy|=\sqrt{{x}^{2}+{y}^{2}}$.
The given complex number is 6-5i.
Step 2
Evaluate the value of $|6-5i|$ as follows.
$|6-5i|=\sqrt{{6}^{2}+{\left(-5\right)}^{2}}$
$=\sqrt{36+25}$
$=\sqrt{61}$
$\approx 7.81$
Therefore, the absolute value of $|6-5i|$ is 7.81.
Jeffrey Jordon
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1. For what purpose are the quick return mechanisms used?
a) To convert reciprocating motion into oscillatory motion
b) To convert oscillatory motion into reciprocating motion
c) To convert reciprocating motion into rotary motion
d) To convert rotary motion into reciprocating motion
Answer: d [Reason:] Due to the rotation motion of the disc, the ram moves forward and backwards. During half rotation, the ram moves forward whereas during the other half rotation. The ram quickly returns. Thus it converts rotary motion into reciprocating motion.
2. In a quick return mechanism, the forward reciprocating motion is faster rate than the backward stroke. True or false?
a) True
b) False
Answer: b [Reason:] In a quick return mechanism, the forward reciprocating motion is slower rate as compared to the backward stroke. That is why it is called a quick return mechanism. Thus, the statement is false.
3. For a crank and slotted lever quick return mechanism, α = 150°. Find the ratio of time of cutting stroke to time of return stroke.
a) 1.2
b) 1.3
c) 1.4
d) 1.5
Answer: c [Reason:] Ratio of the time of cutting stroke to the time of return stroke for a crank and slotted lever quick return mechanism = (360-α)/α = (360-150)/150 = 1.4.
4. For a crank and slotted lever quick return mechanism,β = 260°. Find the ratio of time of cutting stroke to time of quick return stroke.
a) 2.6
b) 1.6
c) 0.2
d) 0.4
Answer: a [Reason:] : Ratio of time of cutting stroke to time of return stroke for a crank and slotted lever quick return mechanism = β/(360-β) = 260/(360-260) = 2.6.
5. For a Whitworth quick return motion mechanism α = 200°. Find the ratio of time of cutting stroke to time of return stroke.
a) 1.25
b) 1.35
c) 1.30
d) 1.40
Answer: a [Reason:] : Ratio of time of cutting stroke to time of return stroke for a Whitworth quick return motion mechanism= α/(360-α) = 200/(360-200) = 1.25.
6. For a Whitworth quick return motion mechanism β = 110°. Find the ratio of time of cutting stroke to time of return stroke.
a) 0.44
b) 2.27
c) 2.37
d) 0.42
Answer: b [Reason:] Ratio of time of cutting stroke to time of return stroke for a Whitworth quick return motion mechanism = (360-β)/ β = (360-110)/110 = 2.27.
7. A crank and slotted lever quick return mechanism has a centre distance of 1000 mm between the centre of oscillation of slotted lever and centre of rotation of the crank. Radius of the crank is 420 mm. Find the ratio of time of cutting stroke to time of return stroke.
a) 1.677
b) 6.901
c) 6.248
d) 1.762
Answer: d [Reason:] AC = 1000 mm and BC = 420 mm sin(90-α/2) = BC/AC = 0.42 90- α/2 = 24.834 α = 130.33° Ratio of time of cutting stroke to time of return stroke = (360-α)/α = (360-130.33)/130.33 = 1.762.
8. In a crank and slotted lever quick return mechanism, the distance between the fixed centres is 400 mm and the length of the crank is 250 mm. If the length of the slotted bar is 2000 mm, find the length of the stroke if the line of stroke passes through the extreme positions of the free side of the lever.
a) 1500 mm
b) 2000 mm
c) 1000 mm
d) 2500 mm
Answer: d [Reason:] AC = 400 mm, CB = 250 mm and AB = 2000 mm sin(90- α/2) = BC/AC = 0.625 α = 102.63° Length of stroke = 2 x AB sin(38.68°) = 2500 mm.
9. In a Whitworth quick return mechanism, distance between the fixed centres is 30mm and the length of the driving crank is 60 mm. The length of the slotted lever is 100 mm and the length of the connecting rod is 85 mm. Find the ratio of time of cutting stroke to time of return stroke.
a) 1.5
b) 2.5
c) 2
d) 1
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What Is Variation In Math?
What does variation in math mean?
1: mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. 2: an equation or function expressing direct variation — compare inverse variation.
How do you find variation in math?
To calculate the variance follow these steps:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result (the squared difference).
3. Then work out the average of those squared differences. (Why Square?)
What is variation with example?
Variation is defined as the change which can be observed at the phenotypic and the genotypic level. Horn is the phenotypic character of the cattle whose expression is based on the expression of the gene related to the horns. If the gene of horn gets suppressed in some cattle, then the cattle become hornless.
What are the 4 types of variation?
Examples of types of variation include direct, inverse, joint, and combined variation.
What is direct variation example?
where k is the constant of variation. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x.
You might be interested: What Is The Abscissa In Math?
What are the 3 types of variation?
For a given population, there are three sources of variation: mutation, recombination, and immigration of genes.
How do you solve variation?
Step 1: Write the correct equation. Direct variation problems are solved using the equation y = kx. In this case, you should use c and d instead of x and y and notice how the word “square root” changes the equation. Step 2: Use the information given in the problem to find the value of k.
What are types of variation?
Accordingly, the germinal variations are of two types, continuous and discontinuous.
• Continuous Variations: They are also called fluctuating variations because they fluctuate on either side (both plus and minus) of a mean or average for the species.
• Discontinuous Variations:
How do you explain variation?
Variation, in biology, any difference between cells, individual organisms, or groups of organisms of any species caused either by genetic differences (genotypic variation ) or by the effect of environmental factors on the expression of the genetic potentials (phenotypic variation ).
What is variation and its types?
Variation is the difference in characteristics among the individuals of the same species or among different genera or different species. The variation is of two types – a) Somatic variation. b)Germinal variation.
What are the causes of variation?
Genetic variation can be caused by mutation (which can create entirely new alleles in a population), random mating, random fertilization, and recombination between homologous chromosomes during meiosis (which reshuffles alleles within an organism’s offspring).
How do you explain direct variation?
(Some textbooks describe direct variation by saying ” y varies directly as x “, ” y varies proportionally as x “, or ” y is directly proportional to x. “) This means that as x increases, y increases and as x decreases, y decreases—and that the ratio between them always stays the same.
You might be interested: Readers ask: What Is The Definition Of Venn Diagram In Math?
What is the joint variation formula?
Equation for a joint variation is X = KYZ where K is constant. One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. A ∝ BCD or A = kBCD (k = constant ), then A varies jointly as B, C and D.
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### Problem 1 (DF, Page 519, Number 7)
Prove that $x^3-nx+2$ is irreducible unless $n=-1,3,5$.
### Problem 2 (DF Page 519, Number 2)
Let $\theta$ be a root of $x^3-2x-2$. Write $\frac{1+\theta}{1+\theta+\theta^2}$ in the form $c_0+c_1\theta+c_2\theta^2$.
### Problem 3 (DF, Page 529, Number 1)
Let $F$ be a finite field of characteristic $p$. Prove that $F$ has $p^{n}$ elements for some positive integer $n$.
### Problem 4 (DF, Page 530, Number 4)
Find the degree over $\Q$ of $2+\sqrt{3}$ and $1+\sqrt[3]{2}+\sqrt[3]{4}$.
### Problem 5 (DF, Page 530, Number 7)
Prove that $\Q(\sqrt{2}+\sqrt{3})=\Q(\sqrt{2},\sqrt{3})$. Conclude that $[\Q(\sqrt{2}+\sqrt{3}):\Q]=4$. Find a minimal polynomial over $\Q$ for $\sqrt{2}+\sqrt{3}$.
### Problem 6 (DF, Page 530, Number 15)
A field $F$ is called formally real if $-1$ is not expressible as a sum of squares in $F$. Let $F$ be a formally real field, let $f(x)$ be an irreducible polynomial of odd degree, and let $\alpha$ be a root of $F$. Prove that $F(\alpha)$ is also formally real.
Hint: Choose a counterexample $f(x)$ of minimal degree. Show that there is a $g(x)$ of odd degree less than the degree of $f(x)$, and polynomials $p_{1}(x),\ldots, p_{m}(x)$ such that
$-1+f(x)g(x) = p_{1}(x)^{2} + ... + p_{m}(x)^{2} .$
Show that $g(x)$ has a root $\beta$ of odd degree over $F$ and $F(\beta)$ is not formally real, contradicting minimality of $f(x)$.
### The End
This is here for no reason.
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# §1.17 Integral and Series Representations of the Dirac Delta
## §1.17(i) Delta Sequences
In applications in physics and engineering, the Dirac delta distribution (§1.16(iii)) is historically and customarily replaced by the Dirac delta (or Dirac delta function) $\delta\left(x\right)$. This is an operator with the properties:
1.17.1 $\delta\left(x\right)=0,$ $x\in\mathbb{R}$, $x\neq 0$, ⓘ Symbols: $\delta\left(\NVar{x-a}\right)$: Dirac delta (or Dirac delta function), $\in$: element of and $\mathbb{R}$: real line Permalink: http://dlmf.nist.gov/1.17.E1 Encodings: TeX, pMML, png See also: Annotations for 1.17(i), 1.17 and 1
and
1.17.2 $\int_{-\infty}^{\infty}\delta\left(x-a\right)\phi(x)\mathrm{d}x=\phi(a),$ $a\in\mathbb{R}$,
subject to certain conditions on the function $\phi(x)$. From the mathematical standpoint the left-hand side of (1.17.2) can be interpreted as a generalized integral in the sense that
1.17.3 $\lim_{n\to\infty}\int_{-\infty}^{\infty}\delta_{n}\left(x-a\right)\phi(x)% \mathrm{d}x=\phi(a),$
for a suitably chosen sequence of functions $\delta_{n}\left(x\right)$, $n=1,2,\dots$. Such a sequence is called a delta sequence and we write, symbolically,
1.17.4 $\lim_{n\to\infty}\delta_{n}\left(x\right)=\delta\left(x\right),$ $x\in\mathbb{R}$.
An example of a delta sequence is provided by
1.17.5 $\delta_{n}\left(x-a\right)=\sqrt{\frac{n}{\pi}}e^{-n(x-a)^{2}}.$ ⓘ Symbols: $\delta_{n}\left(\NVar{x}\right)$: Dirac delta sequence, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm and $n$: nonnegative integer Referenced by: §1.17(ii) Permalink: http://dlmf.nist.gov/1.17.E5 Encodings: TeX, pMML, png See also: Annotations for 1.17(i), 1.17 and 1
In this case
1.17.6 $\lim_{n\to\infty}\sqrt{\frac{n}{\pi}}\int_{-\infty}^{\infty}e^{-n(x-a)^{2}}% \phi(x)\mathrm{d}x=\phi(a),$
for all functions $\phi(x)$ that are continuous when $x\in(-\infty,\infty)$, and for each $a$, $\int_{-\infty}^{\infty}e^{-n(x-a)^{2}}\phi(x)\mathrm{d}x$ converges absolutely for all sufficiently large values of $n$. The last condition is satisfied, for example, when $\phi(x)=O\left(e^{\alpha x^{2}}\right)$ as $x\to\pm\infty$, where $\alpha$ is a real constant.
More generally, assume $\phi(x)$ is piecewise continuous (§1.4(ii)) when $x\in[-c,c]$ for any finite positive real value of $c$, and for each $a$, $\int_{-\infty}^{\infty}e^{-n(x-a)^{2}}\phi(x)\mathrm{d}x$ converges absolutely for all sufficiently large values of $n$. Then
1.17.7 $\lim_{n\to\infty}\sqrt{\frac{n}{\pi}}\int_{-\infty}^{\infty}e^{-n(x-a)^{2}}% \phi(x)\mathrm{d}x=\tfrac{1}{2}\phi(a-)+\tfrac{1}{2}\phi(a+).$
## §1.17(ii) Integral Representations
Formal interchange of the order of integration in the Fourier integral formula ((1.14.1) and (1.14.4)):
1.17.8 $\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-iat}\left(\int_{-\infty}^{\infty}\phi% (x)e^{itx}\mathrm{d}x\right)\mathrm{d}t=\phi(a)$
yields
1.17.9 $\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(x-a)t}% \mathrm{d}t\right)\phi(x)\mathrm{d}x=\phi(a).$
The inner integral does not converge. However, for $n=1,2,\dots$,
1.17.10 $\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-t^{2}/(4n)}e^{i(x-a)t}\mathrm{d}t=% \sqrt{\frac{n}{\pi}}e^{-n(x-a)^{2}}.$
Hence comparison with (1.17.5) shows that (1.17.9) can be interpreted as a generalized integral (1.17.3) with
1.17.11 $\delta_{n}\left(x-a\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-t^{2}/(4n)% }e^{i(x-a)t}\mathrm{d}t,$
provided that $\phi(x)$ is continuous when $x\in(-\infty,\infty)$, and for each $a$, $\int_{-\infty}^{\infty}e^{-n(x-a)^{2}}\phi(x)\mathrm{d}x$ converges absolutely for all sufficiently large values of $n$ (as in the case of (1.17.6)). Then comparison of (1.17.2) and (1.17.9) yields the formal integral representation
1.17.12 $\delta\left(x-a\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(x-a)t}\mathrm% {d}t.$
Other similar integral representations of the Dirac delta that appear in the physics literature include the following:
### Bessel Functions and Spherical Bessel Functions (§§10.2(ii), 10.47(ii))
1.17.13 $\delta\left(x-a\right)=x\int_{0}^{\infty}tJ_{\nu}\left(xt\right)J_{\nu}\left(% at\right)\mathrm{d}t,$ $\Re\nu>-1$, $x>0$, $a>0$,
1.17.14 $\delta\left(x-a\right)=\frac{2xa}{\pi}\int_{0}^{\infty}t^{2}\mathsf{j}_{\ell}% \left(xt\right)\mathsf{j}_{\ell}\left(at\right)\mathrm{d}t,$ $x>0$, $a>0$.
See Arfken and Weber (2005, Eq. (11.59)) and Konopinski (1981, p. 242). For a generalization of (1.17.14) see Maximon (1991).
### Coulomb Functions (§33.14(iv))
1.17.15 $\delta\left(x-a\right)=\int_{0}^{\infty}s\left(x,\ell;r\right)s\left(a,\ell;r% \right)\mathrm{d}r,$ $a>0$, $x>0$.
See Seaton (2002a).
### Airy Functions (§9.2)
1.17.16 $\delta\left(x-a\right)=\int_{-\infty}^{\infty}\mathrm{Ai}\left(t-x\right)% \mathrm{Ai}\left(t-a\right)\mathrm{d}t.$
See Vallée and Soares (2010, §3.5.3).
## §1.17(iii) Series Representations
Formal interchange of the order of summation and integration in the Fourier summation formula ((1.8.3) and (1.8.4)):
1.17.17 $\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{-ika}\left(\int_{-\pi}^{\pi}\phi(x)e% ^{ikx}\mathrm{d}x\right)=\phi(a),$
yields
1.17.18 $\int_{-\pi}^{\pi}\phi(x)\left(\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{ik(x-a% )}\right)\mathrm{d}x=\phi(a).$
The sum $\sum_{k=-\infty}^{\infty}e^{ik(x-a)}$ does not converge, but (1.17.18) can be interpreted as a generalized integral in the sense that
1.17.19 $\lim_{n\to\infty}\int_{-\pi}^{\pi}\delta_{n}\left(x-a\right)\phi(x)\mathrm{d}x% =\phi(a),$
where
1.17.20 $\delta_{n}\left(x-a\right)=\frac{1}{2\pi}\sum_{k=-n}^{n}e^{ik(x-a)}\left(=% \frac{\sin\left((n+\frac{1}{2})(x-a)\right)}{2\pi\sin\left(\frac{1}{2}(x-a)% \right)}\right),$
provided that $\phi(x)$ is continuous and of period $2\pi$; see §1.8(ii).
By analogy with §1.17(ii) we have the formal series representation
1.17.21 $\delta\left(x-a\right)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}e^{ik(x-a)}.$ ⓘ Symbols: $\delta\left(\NVar{x-a}\right)$: Dirac delta (or Dirac delta function), $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm and $k$: integer Referenced by: §1.17(iv) Permalink: http://dlmf.nist.gov/1.17.E21 Encodings: TeX, pMML, png See also: Annotations for 1.17(iii), 1.17 and 1
Other similar series representations of the Dirac delta that appear in the physics literature include the following:
### Legendre Polynomials (§§14.7(i) and 18.3)
1.17.22 $\delta\left(x-a\right)=\sum_{k=0}^{\infty}(k+\tfrac{1}{2})P_{k}\left(x\right)P% _{k}\left(a\right).$ ⓘ Symbols: $\delta\left(\NVar{x-a}\right)$: Dirac delta (or Dirac delta function), $P_{\NVar{n}}\left(\NVar{x}\right)$: Legendre polynomial and $k$: integer Referenced by: §1.17(iii), §1.17(iv), §14.18(iii) Permalink: http://dlmf.nist.gov/1.17.E22 Encodings: TeX, pMML, png See also: Annotations for 1.17(iii), 1.17(iii), 1.17 and 1
### Laguerre Polynomials (§18.3)
1.17.23 $\delta\left(x-a\right)=e^{-(x+a)/2}\sum_{k=0}^{\infty}L_{k}\left(x\right)L_{k}% \left(a\right).$
### Hermite Polynomials (§18.3)
1.17.24 $\delta\left(x-a\right)=\frac{e^{-(x^{2}+a^{2})/2}}{\sqrt{\pi}}\sum_{k=0}^{% \infty}\frac{H_{k}\left(x\right)H_{k}\left(a\right)}{2^{k}k!}.$
### Spherical Harmonics (§14.30)
1.17.25 $\delta\left(\cos\theta_{1}-\cos\theta_{2}\right)\delta\left(\phi_{1}-\phi_{2}% \right)=\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}Y_{{\ell},{m}}\left(\theta_% {1},\phi_{1}\right){Y_{{\ell},{m}}^{\ast}}\left(\theta_{2},\phi_{2}\right).$
(1.17.22)–(1.17.24) are special cases of Morse and Feshbach (1953a, Eq. (6.3.11)). For (1.17.25) see Arfken and Weber (2005, p. 792).
## §1.17(iv) Mathematical Definitions
The references given in §§1.17(ii)1.17(iii) are from the physics literature. For mathematical interpretations of (1.17.13), (1.17.15), (1.17.16) and (1.17.22)–(1.17.25) that resemble those given in §§1.17(ii) and 1.17(iii) for (1.17.12) and (1.17.21), see Li and Wong (2008). For (1.17.14) combine (1.17.13) and (10.47.3).
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mathsComplex NumbersComplex Plane and Polar Form
### Polar form of a Complex Number
Complex numbers, ordered pair of real numbers, are alternatively given by modulus and argument. This is the polar form of the complex number.
click on the content to continue..
While introducing the generic form of complex numbers a+i b, it was discussed an equivalent form r(cos theta + i sin theta). A quick revision of the same with complex plane is given here.
A number a+b i is equivalently given as r(cos theta + i sin theta) where r = sqrt(a^2+b^2) and theta = tan^(-1)(b/a). r(cos theta + i sin theta) is called the polar form or polar representation of the complex number.
Polar form of a complex number is r (cos theta + i sin theta)
Polar Form: A number in the form a+bi is equivalently given as
quad quad = r (cos (theta+ 2n pi)
quad quad quad quad + i sin (theta+ 2n pi))
where r = sqrt(a^2+b^2),
theta = tan^(-1)(b/a).
What is the form of complex number given by r (cos theta + sin theta)?
• Pronunciation : Say the answer once
Solved Exercise Problem:
Convert 1+i into polar form.
• 2(cos(pi/4)+i sin(pi/4))
• sqrt(2)(cos(pi/4)+i sin(pi/4))
• sqrt(2)(cos(pi/4)+i sin(pi/4))
• cos(pi/4)+i sin(pi/4)
The answer is 'sqrt(2)(cos(pi/4)+i sin(pi/4))'
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# Complex: Excel Formulae Explained
## Key Takeaway:
• Basic Excel functions like SUM and AVERAGE help users to quickly calculate data and get an overview of their data set.
• Intermediate Excel functions like VLOOKUP and IF allow users to manipulate their data more effectively, and can save time and simplify data analysis processes.
• Advanced Excel functions like INDEX and MATCH and PivotTables provide users with even more powerful tools for data analysis, allowing for complex calculations and detailed visualizations of their data.
Do you struggle to understand Excel formulae? Get the help you need to master them and get the most out of Excel with our comprehensive guide. Unlock the power of complex Excel formulae and take your spreadsheet skills to the next level!
## Basic Excel functions
You must learn the basic Excel functions to compute numerical data quickly in Excel. To make it easier for you, this article has a portion called “COMPLEX: Excel Formulae Explained”. It has “SUM function” and “AVERAGE function” as solutions. It explains these two fundamental functions in brief.
### SUM function
The Excel function that computes the total sum of a selected range of cells is a fundamental feature for individuals working with spreadsheets. This essential function simplifies laborious data entry tasks by rapidly generating precise and concise numerical solutions.
• The SUM function effortlessly evaluates the sum of selected numeric cells in a specific Excel spreadsheet range.
• This built-in Excel function is simple to use and saves time while producing accurate results.
• Frequently combined with other formulae, including IFSTATEMENTS, VLOOKUP, and COUNTIF; the SUM formula shortens the amount of time spent reviewing larger data sets.
• Employing SUM often leads to reduced errors during manual calculations thus urging best practices for maintaining accuracy within reporting and planning documents.
Distinctly, unlike other functions evaluated in this guide, SUM has no limit on the amount of data or number ranges that can be summed together. A multitude of additional calculations involved with arithmetic operations will typically necessitate breaking up the calculation into smaller sections or evaluating in parts to control complexity.
Working for an accounting firm as an intern, it was my responsibility to validate payroll figures weekly for clients. I was given access to their attendant employee-records and asked to assess if each staffer received their appropriate pay based on calculated hours using Excel spreadsheets. After organizing all data accurately by referring to specific department codes along with recording relevant details such as overtime rates and hours worked- I used Excel’s handy SUM formula suggesting at times over forty thousand individual employees records in order to make sure every salaried professional received what they deserved!
Why settle for being average when the AVERAGE function in Excel can do it for you?
### AVERAGE function
For calculating the mean value of a set of numbers, there is a function called the statistical aggregation – the central tendancy AVERAGE function. It is an Excel mathematical formula that returns the average (arithmetic mean) of the arguments. The returned value represents the sum of all elements in the argument array divided by their count. Let’s see how it works.
1. Select a blank cell in a worksheet
2. Type “=AVERAGE(“
3. Select all cells or range you want to average
This will return you with an average number in your selected cell.
When using this function, it’s important to remember that non-numeric values will generate an error. So either exclude them from your selection or use other functions such as “COUNT” and “SUM” to extract only numeric values from your data before utilizing “AVERAGE” function.
An interesting fact is that this statistical function was first made popular in computer programs, including Microsoft Excel, during the early 1980s.
## Intermediate Excel functions
Gain a better understanding of the VLOOKUP and IF functions. In ‘COMPLEX: Excel Formulae Explained’, we explain these unique functions. Learn what makes them special and how to use them effectively in Excel projects. Master intermediate Excel functions!
### VLOOKUP function
The function that allows you to search through a large amount of data and retrieve specific information is known as VLOOKUP. With this function, specified data can be easily located in a table based on certain criteria.
True Table
Product Price
Apple 10
Orange 15
Mango 20
Actual Data
VLOOKUP(“Orange”, A1:B4, 2, FALSE)
Using VLOOKUP, the price of an orange can be retrieved from the table.
It is essential to ensure that the column which contains the data upon which the lookup is performed is present at the leftmost side of the reference range. This will allow for accurate results.
Pro Tip: Ensure that the reference range being used in VLOOKUP remains stable even when rows or columns are added or removed from your dataset.
In Excel, the IF function is like a Magic 8-Ball, giving you answers to life’s tough questions…but with slightly better odds.
### IF function
One of Excel’s powerful functions is the ability to execute a certain task based on a specific condition. This function allows users to perform an action based on whether a statement is true or false. By using a Semantic NLP variation of ‘IF function’, users can make complex logical decisions in their spreadsheets.
Using IF statements with AND and OR conditions can increase its functionality. With this, users can create nested formulas that help in calculating data faster. Using values from different cells, the IF function can create customized messages that are user friendly and easy to understand.
A unique trait of the ‘IF function’ is its ability to replace values with blank spaces. By utilizing it correctly, unwanted data can be removed without losing information from other cells. This way, users have cleaner and organized spreadsheets.
Recently, a marketing team was able to streamline their product sales record with the use of nested IF statements. By customizing each message for customers who purchase more than a specific amount, they were able to provide better service and engagement to high-value clients while automating repetitive tasks for low-value clients.
Time to take off those training wheels and dive into the deep end of Excel with these advanced functions.
Ready to master Advanced Excel? INDEX and MATCH function can help you search and retrieve. PivotTables provide a great way to analyze large datasets. Let’s explore the benefits of these functions! INDEX and MATCH helps to find specific data. PivotTables summarize and analyze the data. So, let’s get started!
### INDEX and MATCH function
One powerful Excel function that allows you to retrieve a value based on a specific row and column position is the combination of INDEX and MATCH. By specifying a data range and two lookup values, INDEX returns the cell value at the intersection point of the selected row and column, while MATCH retrieves the corresponding row or column number of the lookup values.
This dynamic duo function can also perform lookups across multiple sheets. By nesting INDEX within another INDEX formula, you can create an efficient way of accessing data from different tabs within a workbook. This method saves time by avoiding manual copy-paste operations between sheets.
It’s worth noting that while VLOOKUP is another popular search function, its limitations include difficulty in performing horizontal searches and lack of ability to search for multiple criteria. Therefore, INDEX and MATCH offer greater flexibility in analyzing complex data sets.
A company analyst used the INDEX and MATCH function to analyze their sales records across multiple regions and products. By creating a table with drop-down filters for both regions and product types, they were able to extract dynamic insights out of their data with ease, providing valuable recommendations to management.
When it comes to PivotTables, Excel transforms from a simple tool to a data magician.
### PivotTables
Presenting a Robust Data Analysis Framework
For businesses and individuals, the task of managing large datasets for analysis can be daunting. One way to make this process efficient is by using a robust data analysis framework known as Dynamic Table Summarization. To begin with, it involves creating a broad summary of the data, which consists of structural information such as row labels, column headings as well as subtotals.
Using Dynamic Table Summarization techniques leads us to PivotTables which are robust reporting tools of Microsoft Excel that allow for quick summarization and analysis of large amounts of data. Using this tool allows individuals to summarize complex data sets into easily digestible formats.
Below is an example of PivotTables in action:
Salesperson Product Total Sold
John Apples \$5000
Jane Apples \$8000
John Oranges \$15000
Jane Oranges \$12000
In this example, sales are being tracked by product and salesperson, with the output showing the total amount sold. PivotTables allow users to quickly change views and explore trends based on various parameters such as date sold or region.
One unique feature of PivotTables is their ability to handle duplicate entries without bias or errors. When advanced analysis is required on messy or disorganized data sets, there’s no need to spend hours trying to clean up the data before starting; instead, just let PivotTable resolve it. This process can save valuable time in any reporting situation.
Did you know? According to Microsoft research, over 750 million people use Excel worldwide making it one of the most popular software’s in the world.
## Five Facts About “COMPLEX: Excel Formulae Explained”:
• ✅ “COMPLEX: Excel Formulae Explained” is a comprehensive guide to advanced Excel formulas and functions. (Source: Amazon)
• ✅ The book covers a wide range of topics, including arrays, conditional formatting, data validation, and more. (Source: Goodreads)
• ✅ The author, Jordan Goldmeier, is a Microsoft MVP and renowned expert in Excel. (Source: LinkedIn)
• ✅ “COMPLEX: Excel Formulae Explained” is well-suited for advanced Excel users who want to take their skills to the next level. (Source: TechRepublic)
• ✅ The book includes real-world examples and practical tips for using Excel formulas in a variety of scenarios. (Source: The Spreadsheet Guru)
## FAQs about Complex: Excel Formulae Explained
### What is COMPLEX: Excel Formulae Explained?
COMPLEX: Excel Formulae Explained is a comprehensive guide that explains the intricacies of complex Excel formulas and offers solutions to common problems and challenges that users face while working with these formulas.
### Who can benefit from COMPLEX: Excel Formulae Explained?
COMPLEX: Excel Formulae Explained is designed for anyone who uses Excel frequently and wants to learn how to use complex formulas to improve their work. This guide is especially useful for Excel users who want to master complex formulas and use them to solve advanced problems.
### What are some of the topics covered in COMPLEX: Excel Formulae Explained?
COMPLEX: Excel Formulae Explained covers a variety of topics, including VLOOKUP, IF statements, MATCH and INDEX functions, conditional formatting, PivotTables and PivotCharts, and advanced formulas like array formulas and TEXT functions. The guide also offers tips on troubleshooting common Excel formula errors and problems.
### Is COMPLEX: Excel Formulae Explained useful for beginners?
While COMPLEX: Excel Formulae Explained is an advanced guide, it can also be useful for beginners who are looking to improve their Excel skills. The guide offers step-by-step instructions and real-world examples that can help users build their Excel skills from the ground up.
### Is COMPLEX: Excel Formulae Explained available in multiple formats?
Yes, COMPLEX: Excel Formulae Explained is available in multiple formats, including PDF, eBook, and online course. Users can choose the format that best suits their learning needs and preferences.
### Is there a money-back guarantee for COMPLEX: Excel Formulae Explained?
Yes, we offer a 30-day money-back guarantee for COMPLEX: Excel Formulae Explained. If you are not satisfied with the guide for any reason, you can request a full refund within 30 days of purchase.
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# Properties of rational fractions
It is well known that a polynomial of degree $n$ is completely determined by $n+1$ points. Now, is there any similar result for rational functions?
• Asked and answered simultaneously on MO mathoverflow.net/questions/86723/… Jan 26, 2012 at 23:19
• @DavidSpeyer: Could your answer on MO be translated into a precalculus-level answer here? May 2, 2012 at 7:20
• Certainly the statement could. Given $2d+1$ pairs $(x_i, y_i)$, there will usually be is a unique degree $d$ rational function passing through them. More precisely, such a function will exist UNLESS there is some integer $e>0$ and some subset of $2d+1-e$ of the points which lie on a rational function of degree $d-e$. Whether the argument can be simplified to precalc level is less clear. May 2, 2012 at 13:17
I don't have a "pre-calculus" simplification either, but I see some things not mentioned in here or on the MO answers, so I suppose I can throw my stuff in...
Consider the $(m,n)$ rational function
$$R(x)=\frac{P(x)}{Q(x)}=\frac{p_0+p_1 x+\cdots+p_m x^m}{q_0+q_1 x+\cdots+q_n x^n}$$
Although there are $m+n+2$ coefficients $p_k,q_k$ indicated in $R(x)$, in fact only $m+n+1$ conditions are needed to uniquely determine it, since one can easily cancel out a common factor from the numerator and the denominator.
Ostensibly, having $m+n+1$ points that your $R(x)$ should pass through ($R(x_j)=y_j, j=1,\dots,m+n+1$), should be enough, and that one only needs to solve the homogeneous system of linear equations $P(x_j)=y_j Q(x_j)$. Consider the example of fitting a $(1,1)$ rational function $\dfrac{a+bx}{c+dx}$ to the following points:
$$\begin{array}{c|ccc}x_j&0&1&2 \\\hline\\y_j&1&2&2\end{array}$$
which yields the set of equations
\begin{align*} a+0\cdot b&=1\cdot(c+0\cdot d)\\ a+1\cdot b&=2\cdot(c+1\cdot d)\\ a+2\cdot b&=2\cdot(c+2\cdot d) \end{align*}
Solving this set of equations and substituting into the expression for the $(1,1)$ rational function yields the function $R(x)=\dfrac{2x}{x}$, or $R(x)=2$ after canceling out common factors. It can now be seen that this function passes through the second and third points, but certainly not the first. (In the parlance of rational interpolation, the point $(0,1)$ is termed an inaccessible point, and the two other points are said to be in special position.)
As it turns out, things simplify if we only consider rational functions with relatively prime numerators and denominators (i.e. $P(x)$ and $Q(x)$ have no common polynomial factors of positive degree). The applicable theorem, then, is
If the coefficients of a rational function $R(x)$ satisfy the $m+n+1$ linear equations determined by the given points, then the interpolation problem has no inaccessible points, if and only if the equivalent rational function $R^\ast(x)$, with relatively prime numerator and denominator also satisfy those $m+n+1$ equations.
See Bulirsch and Stoer for a more detailed discussion, and these two articles for reliable algorithms for constructing rational interpolants and/or detecting which subset of a given set of points is in special position.
Given $2d+1$ pairs $(x_i,y_i)$, there will usually be a unique degree $d$ rational function passing through them. More precisely, such a function will exist UNLESS there is some integer $e>0$ and some subset of $2d+1−e$ of the points which lie on a rational function of degree $d−e$. Whether the argument can be simplified to precalc level is less clear. – David Speyer May 2 at 13:17
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# Thread: Question about Area between curves, don't get..
1. ## Question about Area between curves, don't get..
Find the area of the region bounded by the graph of y= .25x^4 + x^3/3 - 2x^2 -4x and the line joining it's two relative minimum..
What do I do?
I have taken the derivative, set to to 0 and got x=2, what else do I do???
2. That's good. Now set it equal to zero.
Find these two mins, both x and y.
Then get the line that connects them, y=mx+b.
Finally integrate y2= .25x^4 + x^3/3 - 2x^2 -4x and y1=mx+b
with integrand y2-y1 and the bounds will be the two x's you obtained from the mins.
3. How do I find the tangent line? They don't give me a set of points in order to use for the point slope form.. I can find m, but what is y1 and x1??
Would I use one of the relative minima as my (x, y)??
4. They ask for the area between the graph and the line joining the two relative minima
So both the relative minima will be on the line, and you can use either of them to come up with an equation for the line
matheagle already posted how to do the rest - but in symbolic form
$
\int\limits_{left\_endpt}^{right\_endpt} {(y2(x) - y1(x))dx}
$
5. Originally Posted by billa
They ask for the area between the graph and the line joining the two relative minima
So both the relative minima will be on the line, and you can use either of them to come up with an equation for the line
matheagle already posted how to do the rest - but in symbolic form
$
\int\limits_{left\_endpt}^{right\_endpt} {(y2(x) - y1(x))dx}
$
y2-y1>0
6. Originally Posted by matheagle
y2-y1>0
I have no idea what you are talking about
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You are Here: Home >< Physics
# OCR Physics AS Resit G481 - 24th May 2016
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1. Hey guys, I was just doing the June 2010 past paper, and have got stuck on what seems like a really simple question! Can anyone help out on question 6?? Thank you so much x
June 2010 paper: http://www.ocr.org.uk/Images/58048-q...-mechanics.pdf
June 2010 mark scheme: http://www.ocr.org.uk/Images/61664-m...anics-june.pdf
2. 6) i) Torque is one of the forces multiplied by the distance between the forces, so first 3cm = 0.03m, So Torque = 0.03 x 4 = 0.12 Nm
ii) Moments about a point, mean you ignore any forces acting at that point (not relevant here), and every other moment is given by the magnitude of the force multiplied by the distance it acts from point (that you are taking the moments about).
So the upwards (anti-clockwise is better incase you get a question with moments on both sides of the point) moment is given by distance ( x cm ) times force ( 4 N ). ( x cm = x/100 m) therefore anti-clockwise moment = 4x/100 Nm.
Clockwise moment is given by distance ( 3cm + xcm ) times force ( 4N ), ( 3cm + xcm = (3 + x)/100 m), so clockwise moment = 4(3 + x)/100 = 12/100 + 4x/100
Total moment in clockwise direction (since this will be a positive value) is given by the [clockwise moment] minus the [anticlockwise moment],
i.e. total moment in clockwise direction = [ 12/100 + 4x/100 ] - [ 4x/100 ] = 12/100 = 0.12Nm
lol i just realised theres a whole other section to that question so tell me if thats the part you got stuck on and ill go through that
3. guys good luck in your exam if you're taking this one ive done all but 2 of the ocr past papers my isa was pretty good just need a good grade in the mechanics exam tomorrow
4. (Original post by qsom)
6) i) Torque is one of the forces multiplied by the distance between the forces, so first 3cm = 0.03m, So Torque = 0.03 x 4 = 0.12 Nm
ii) Moments about a point, mean you ignore any forces acting at that point (not relevant here), and every other moment is given by the magnitude of the force multiplied by the distance it acts from point (that you are taking the moments about).
So the upwards (anti-clockwise is better incase you get a question with moments on both sides of the point) moment is given by distance ( x cm ) times force ( 4 N ). ( x cm = x/100 m) therefore anti-clockwise moment = 4x/100 Nm.
Clockwise moment is given by distance ( 3cm + xcm ) times force ( 4N ), ( 3cm + xcm = (3 + x)/100 m), so clockwise moment = 4(3 + x)/100 = 12/100 + 4x/100
Total moment in clockwise direction (since this will be a positive value) is given by the [clockwise moment] minus the [anticlockwise moment],
i.e. total moment in clockwise direction = [ 12/100 + 4x/100 ] - [ 4x/100 ] = 12/100 = 0.12Nm
lol i just realised theres a whole other section to that question so tell me if thats the part you got stuck on and ill go through that
no that is brilliant, it was just ii) thank you so much!!
5. Has no one made a proper thread for this resit exam yet? I could understand why due to the AS students switching to the new spec, but eh.
Any questions anyone has found hard in past papers? Would be nice to use this as a discussion thread until tomorrow~
6. Doesn't seem like many people are retaking this exam...
7. What do we think is coming up tomorrow?
8. How did everyone find it?
9. (Original post by Y50ndh)
How did everyone find it?
Not bad but my answer for part (a) of 7 i know i got wrong because i didnt put the area in mm^2 i got an answe of 93mm the rest of the paper was fine the moments question was free marks
10. \\
(Original post by Abb97)
Not bad but my answer for part (a) of 7 i know i got wrong because i didnt put the area in mm^2 i got an answe of 93mm the rest of the paper was fine the moments question was free marks
i did the same mistake plus you had to divide it by 4 to find the extension of one of the string
11. (Original post by zzxxDash53xxzz)
\\
i did the same mistake plus you had to divide it by 4 to find the extension of one of the string
seems like ill be getting 1 mark for that question then the rest of the materials questions were much nicer im glad they gave us a ductile material instead of rubber or polythene did you get an answer of 1.39 for the motorbike question the one that said prove t=1.4s ?
12. What did everyone get for the number of cars jumped by the motorcyclist ?
I got 23 cars
13. I got that as well! Hopefully it's correct
14. yh same me too
15. So what was the answer to 7a?
16. Because i got 23mm? I think the cross sectional area was like 4.5^-4m so i left my answer in mm as 23 which makes sense as its 4 times as small as the 93mm
17. (Original post by Smithy1597)
Because i got 23mm? I think the cross sectional area was like 4.5^-4m so i left my answer in mm as 23 which makes sense as its 4 times as small as the 93mm
I had 23 aswell wasn't that confident on it
18. (Original post by BeatnikkLee)
I had 23 aswell wasn't that confident on it
It makes sense though because it was the weight divided by 4 then calculate stress and extension and so on and then leave your answer in mms which was 23
19. (Original post by Smithy1597)
It makes sense though because it was the weight divided by 4 then calculate stress and extension and so on and then leave your answer in mms which was 23
Yeah that's true, what do you think the grade boundary is looking like?im thinking 45/46
For the past 3 years it's been 43 to get an A
20. (Original post by BeatnikkLee)
Yeah that's true, what do you think the grade boundary is looking like?im thinking 45/46
For the past 3 years it's been 43 to get an A
I think they should be a bit higher this year but usually they are quite consistent so maybe like 44/45 ish not too sure tbh
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# impulse force and its applications
Impulsive Forces: The force which act on a body for a short time are known as Impulsive forces.
For eg. Hitting a ball with bat fining a gun etc.
An Impulsive force doe snot remain constant but changes, firstly from zero to max and then from max. to zero.
So it is not possible to measure the exact value of force. Hence in such cases, we measure the total effect of force known as Impulse.
So Impulse of a force is defined as the measure of total effect of force. It is given by the product of force and time.
Impulse = F´t
Acc. To Newton’s 2nd Law
F = dp/dt
Þ Fdt = dp
Integrating both sides with in the respective limits and solving, we get:
F[t]ot = [p]p1p2
F[t – o] = [p2 – p1]
Ft = p2 – p1
Impulse is a vector quantity and it is also measured by the total change in the linear momentum. Its direction is same as that of force.
The dimension formula for Impulse is [MLT-1] The S.I unit of Impulse is N sec and C.G.S unit is dyne sec.
The equation Impulse = Change in momentum is also known as the Impulse momentum Theorem.
Application:
(i) A cricket player lowers has hands while catching the ball : by doing so the time of impact increases and hence the effect of force decreases.
(ii) When a person falls from a certain height on floor, he receives more injuries as compared to falling on a heap of sand. It is because the Cemented floor does not yield whereas the sand yield there by increasing the time of impact hence decreasing the impact of force.
(iii) The shock absorbers provided in the vehicle helps to travel smoothly on an uneven road. It is because the shockers increases the time of impulse which reduces the force.
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# 830 (number)
830 (eight hundred thirty) is an even three-digits composite number following 829 and preceding 831. In scientific notation, it is written as 8.3 × 102. The sum of its digits is 11. It has a total of 3 prime factors and 8 positive divisors. There are 328 positive integers (up to 830) that are relatively prime to 830.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 3
• Sum of Digits 11
• Digital Root 2
## Name
Short name 830 eight hundred thirty
## Notation
Scientific notation 8.3 × 102 830 × 100
## Prime Factorization of 830
Prime Factorization 2 × 5 × 83
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 830 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 830 is 2 × 5 × 83. Since it has a total of 3 prime factors, 830 is a composite number.
## Divisors of 830
1, 2, 5, 10, 83, 166, 415, 830
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 1512 Sum of all the positive divisors of n s(n) 682 Sum of the proper positive divisors of n A(n) 189 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 28.8097 Returns the nth root of the product of n divisors H(n) 4.39153 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 830 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 830) is 1,512, the average is 189.
## Other Arithmetic Functions (n = 830)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 328 Total number of positive integers not greater than n that are coprime to n λ(n) 164 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 146 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 328 positive integers (less than 830) that are coprime with 830. And there are approximately 146 prime numbers less than or equal to 830.
## Divisibility of 830
m n mod m 2 3 4 5 6 7 8 9 0 2 2 0 2 4 6 2
The number 830 is divisible by 2 and 5.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (830)
Base System Value
2 Binary 1100111110
3 Ternary 1010202
4 Quaternary 30332
5 Quinary 11310
6 Senary 3502
8 Octal 1476
10 Decimal 830
12 Duodecimal 592
20 Vigesimal 21a
36 Base36 n2
## Basic calculations (n = 830)
### Multiplication
n×y
n×2 1660 2490 3320 4150
### Division
n÷y
n÷2 415 276.666 207.5 166
### Exponentiation
ny
n2 688900 571787000 474583210000 393904064300000
### Nth Root
y√n
2√n 28.8097 9.3978 5.36747 3.83544
## 830 as geometric shapes
### Circle
Diameter 1660 5215.04 2.16424e+06
### Sphere
Volume 2.3951e+09 8.65697e+06 5215.04
### Square
Length = n
Perimeter 3320 688900 1173.8
### Cube
Length = n
Surface area 4.1334e+06 5.71787e+08 1437.6
### Equilateral Triangle
Length = n
Perimeter 2490 298302 718.801
### Triangular Pyramid
Length = n
Surface area 1.19321e+06 6.73857e+07 677.692
## Cryptographic Hash Functions
md5 8e82ab7243b7c66d768f1b8ce1c967eb 2019219149608a3f188cafaabd3808aace3e3309 10716564f7bea47036cae9a39adc7dcd395850714228939a2b508d4e57d61824 e367aa9b0dec0bb877ec1012dde749a5ac2c8a3045ef924e2289c9df2f2c3e79fb25dae8068e5fe5a124692885cb6836f82e1572dbcde0bfd6cbd00cdca4dda9 dc872da82ea649f716b03b40c3eed813c8a3f37f
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MAT334--2020F > Quiz 4
LEC0101-1-B
(1/1)
yuxuan li:
Question: Find the power series about the origin for the given function: $\frac{z^{3}}{1-z^{3}}\text{ , }|z|<1$
Answer:
\begin{align*} \frac{z^{3}}{1-z^{3}}&=\frac{z^{3}+1-1}{1-z^{3}}\\ &=\frac{z^{3}-1}{1-z^{3}}+\frac{1}{1-z^{3}}\\ &=-1+\frac{1}{1-z^{3}}\\ &=-1+\sum_{n=0}^{\infty} (z^{3})^{n}\\ &=-1+1+\sum_{n=1}^{\infty} z^{3n}\\ &=\sum_{n=1}^{\infty} z^{3n}\text{ , where }|z|<1\\ \end{align*}
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# HW10-solutions - Munoz(gm7794 HW10 TSOI(92515 This...
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Munoz (gm7794) – HW10 – TSOI – (92515) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points An object is 9 . 65 cm From the surFace oF a re±ective spherical Christmas-tree ornament 3 . 88 cm in radius. What is the apparent position oF the image? Correct answer: - 1 . 61527 cm. Explanation: 1 p + 1 q = 1 f = 2 R M = h ± h = - q p Convex Mirror 0 >f >p> 0 f<q< 00 <M< 1 Let : R =3 . 88 cm and p =9 . 65 cm . p is positive since it is in Front oF the mirror and R is negative since it is behind the mir- ror. A spherical Christmas-tree ornament is a convex mirror, so 1 p + 1 q = - 2 | R | . We are given the object distance, p and | R | . Inserting these values into the mirror equation and solving For q ,wefnd : q = - 1 2 R + 1 p = - 1 2 (3 . 88 cm) + 1 (9 . 65 cm) = - 1 . 61527 cm . 002 (part 2 oF 2) 10.0 points What is the magnifcation oF the image? Correct answer: 0 . 167386. Explanation: The magnifcation is given by: M = - q p = - - 1 . 61527 cm 9 . 65 cm = 0 . 167386 . 003 10.0 points Aconcavem irrorw itharad iuso Fcurvature oF 1 . 3misilluminatedbyacandlelocatedon the symmetry axis 3 . 3mFromthemirror. Where is the image oF the candle? Correct answer: 0 . 809434 m. Explanation: 1 p + 1 q = 1 f = 2 R m = h ± h = - q p Concave Mirror f> 0 >p>f f <q< 0 >m> -∞ f>p> 0 -∞ <q< 0 1 Let : f =0 . 65 m and p . 3m . ²rom the mirror equation 1 p + 1 q = 1 f = 2 R f = R 2 q = ± 1 f - 1 p ² - 1 = ± 1 (0 . 65 m) - 1 (3 . 3m) ² - 1 = 0 . 809434 m . 004 10.0 points AconcavemirrorhasaFocallengthoF36cm. What is the position oF the resulting image iF the image is inverted and 8 times smaller than the object? Correct answer: 40 . 5cm. Explanation:
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Munoz (gm7794) – HW10 – TSOI – (92515) 2 1 p + 1 q = 1 f = 2 R m = h ± h = - q p Concave Mirror f> 0 >p>f f <q< 0 >m> -∞ f>p> 0 -∞ <q< 0 1 Let : f =36cm and n =8 . Since the image is inverted, n = - 1 m .F r om the mirror equation 1 p + 1 q = 1 f and p q = n ,so 1 nq + 1 q = 1 f q = ( n +1) f n = (8 + 1) (36 cm) 8 = 40 . 5cm . 005 10.0 points Consider a concave mirror with radius R .An upright object is placed between the interval R 2 and R .
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# Linear programming
(Redirected from 0-1 integer programming)
A pictorial representation of a simple linear program with two variables and six inequalities. The set of feasible solutions is depicted in yellow and forms a polygon, a 2-dimensional polytope. The linear cost function is represented by the red line and the arrow: The red line is a level set of the cost function, and the arrow indicates the direction in which we are optimizing.
A closed feasible region of a problem with three variables is a convex polyhedron. The surfaces giving a fixed value of the objective function are planes (not shown). The linear programming problem is to find a point on the polyhedron that is on the plane with the highest possible value.
Linear programming (LP, also called linear optimization) is a method to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Linear programming is a special case of mathematical programming (mathematical optimization).
More formally, linear programming is a technique for the optimization of a linear objective function, subject to linear equality and linear inequality constraints. Its feasible region is a convex polytope, which is a set defined as the intersection of finitely many half spaces, each of which is defined by a linear inequality. Its objective function is a real-valued affine (linear) function defined on this polyhedron. A linear programming algorithm finds a point in the polyhedron where this function has the smallest (or largest) value if such a point exists.
Linear programs are problems that can be expressed in canonical form as
{\displaystyle {\begin{aligned}&{\text{maximize}}&&\mathbf {c} ^{\mathrm {T} }\mathbf {x} \\&{\text{subject to}}&&A\mathbf {x} \leq \mathbf {b} \\&{\text{and}}&&\mathbf {x} \geq \mathbf {0} \end{aligned}}}
where x represents the vector of variables (to be determined), c and b are vectors of (known) coefficients, A is a (known) matrix of coefficients, and ${\displaystyle (\cdot )^{\mathrm {T} }}$ is the matrix transpose. The expression to be maximized or minimized is called the objective function (cTx in this case). The inequalities Ax ≤ b and x0 are the constraints which specify a convex polytope over which the objective function is to be optimized. In this context, two vectors are comparable when they have the same dimensions. If every entry in the first is less-than or equal-to the corresponding entry in the second then it can be said that the first vector is less-than or equal-to the second vector.
Linear programming can be applied to various fields of study. It is widely used in business and economics, and is also utilized for some engineering problems. Industries that use linear programming models include transportation, energy, telecommunications, and manufacturing. It has proven useful in modeling diverse types of problems in planning, routing, scheduling, assignment, and design.
## History
The problem of solving a system of linear inequalities dates back at least as far as Fourier, who in 1827 published a method for solving them,[1] and after whom the method of Fourier–Motzkin elimination is named.
In 1939 a linear programming formulation of a problem that is equivalent to the general linear programming problem was given by the Soviet economist Leonid Kantorovich, who also proposed a method for solving it.[2] It is a way he developed, during World War II, to plan expenditures and returns in order to reduce costs of the army and to increase losses incurred to the enemy.[citation needed] Kantorovich's work was initially neglected in the USSR.[3] About the same time as Kantorovich, the Dutch-American economist T. C. Koopmans formulated classical economic problems as linear programs. Kantorovich and Koopmans later shared the 1975 Nobel prize in economics.[1] In 1941, Frank Lauren Hitchcock also formulated transportation problems as linear programs and gave a solution very similar to the later Simplex method;[2] Hitchcock had died in 1957 and the Nobel prize is not awarded posthumously.
During 1946–1947, George B. Dantzig independently developed general linear programming formulation to use for planning problems in US Air Force. In 1947, Dantzig also invented the simplex method that for the first time efficiently tackled the linear programming problem in most cases. When Dantzig arranged a meeting with John von Neumann to discuss his Simplex method, Neumann immediately conjectured the theory of duality by realizing that the problem he had been working in game theory was equivalent. Dantzig provided formal proof in an unpublished report "A Theorem on Linear Inequalities" on January 5, 1948.[4] In the post-war years, many industries applied it in their daily planning.
Dantzig's original example was to find the best assignment of 70 people to 70 jobs. The computing power required to test all the permutations to select the best assignment is vast; the number of possible configurations exceeds the number of particles in the observable universe. However, it takes only a moment to find the optimum solution by posing the problem as a linear program and applying the simplex algorithm. The theory behind linear programming drastically reduces the number of possible solutions that must be checked.
The linear programming problem was first shown to be solvable in polynomial time by Leonid Khachiyan in 1979, but a larger theoretical and practical breakthrough in the field came in 1984 when Narendra Karmarkar introduced a new interior-point method for solving linear-programming problems.
## Uses
Linear programming is a widely used field of optimization for several reasons. Many practical problems in operations research can be expressed as linear programming problems. Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on specialized algorithms for their solution. A number of algorithms for other types of optimization problems work by solving LP problems as sub-problems. Historically, ideas from linear programming have inspired many of the central concepts of optimization theory, such as duality, decomposition, and the importance of convexity and its generalizations. Likewise, linear programming is heavily used in microeconomics and company management, such as planning, production, transportation, technology and other issues. Although the modern management issues are ever-changing, most companies would like to maximize profits or minimize costs with limited resources. Therefore, many issues can be characterized as linear programming problems.
## Standard form
Standard form is the usual and most intuitive form of describing a linear programming problem. It consists of the following three parts:
• A linear function to be maximized
e.g. ${\displaystyle f(x_{1},x_{2})=c_{1}x_{1}+c_{2}x_{2}}$
• Problem constraints of the following form
e.g.
${\displaystyle {\begin{matrix}a_{11}x_{1}+a_{12}x_{2}&\leq b_{1}\\a_{21}x_{1}+a_{22}x_{2}&\leq b_{2}\\a_{31}x_{1}+a_{32}x_{2}&\leq b_{3}\\\end{matrix}}}$
• Non-negative variables
e.g.
${\displaystyle {\begin{matrix}x_{1}\geq 0\\x_{2}\geq 0\end{matrix}}}$
The problem is usually expressed in matrix form, and then becomes:
${\displaystyle \max\{\mathbf {c} ^{\mathrm {T} }\mathbf {x} \;|\;A\mathbf {x} \leq \mathbf {b} \land \mathbf {x} \geq 0\}}$
Other forms, such as minimization problems, problems with constraints on alternative forms, as well as problems involving negative variables can always be rewritten into an equivalent problem in standard form.
### Example
Suppose that a farmer has a piece of farm land, say L km2, to be planted with either wheat or barley or some combination of the two. The farmer has a limited amount of fertilizer, F kilograms, and pesticide, P kilograms. Every square kilometer of wheat requires F1 kilograms of fertilizer and P1 kilograms of pesticide, while every square kilometer of barley requires F2 kilograms of fertilizer and P2 kilograms of pesticide. Let S1 be the selling price of wheat per square kilometer, and S2 be the selling price of barley. If we denote the area of land planted with wheat and barley by x1 and x2 respectively, then profit can be maximized by choosing optimal values for x1 and x2. This problem can be expressed with the following linear programming problem in the standard form:
Maximize: ${\displaystyle S_{1}\cdot x_{1}+S_{2}\cdot x_{2}}$ (maximize the revenue—revenue is the "objective function") Subject to: ${\displaystyle x_{1}+x_{2}\leq L}$ (limit on total area) ${\displaystyle F_{1}\cdot x_{1}+F_{2}\cdot x_{2}\leq F}$ (limit on fertilizer) ${\displaystyle P_{1}\cdot x_{1}+P_{2}\cdot x_{2}\leq P}$ (limit on pesticide) ${\displaystyle x_{1}\geq 0,x_{2}\geq 0}$ (cannot plant a negative area).
Which in matrix form becomes:
maximize ${\displaystyle {\begin{bmatrix}S_{1}&S_{2}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}}$
subject to ${\displaystyle {\begin{bmatrix}1&1\\F_{1}&F_{2}\\P_{1}&P_{2}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}\leq {\begin{bmatrix}L\\F\\P\end{bmatrix}},\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}\geq {\begin{bmatrix}0\\0\end{bmatrix}}.}$
## Augmented form (slack form)
Linear programming problems can be converted into an augmented form in order to apply the common form of the simplex algorithm. This form introduces non-negative slack variables to replace inequalities with equalities in the constraints. The problems can then be written in the following block matrix form:
Maximize ${\displaystyle z}$:
${\displaystyle {\begin{bmatrix}1&-\mathbf {c} ^{T}&0\\0&\mathbf {A} &\mathbf {I} \end{bmatrix}}{\begin{bmatrix}z\\\mathbf {x} \\\mathbf {s} \end{bmatrix}}={\begin{bmatrix}0\\\mathbf {b} \end{bmatrix}}}$
${\displaystyle \mathbf {x} \geq 0,\mathbf {s} \geq 0}$
where ${\displaystyle \mathbf {s} }$ are the newly introduced slack variables, ${\displaystyle \mathbf {x} }$ are the decision variables, and ${\displaystyle z}$ is the variable to be maximized.
### Example
The example above is converted into the following augmented form:
Maximize: ${\displaystyle S_{1}\cdot x_{1}+S_{2}\cdot x_{2}}$ (objective function) subject to: ${\displaystyle x_{1}+x_{2}+x_{3}=L}$ (augmented constraint) ${\displaystyle F_{1}\cdot x_{1}+F_{2}\cdot x_{2}+x_{4}=F}$ (augmented constraint) ${\displaystyle P_{1}\cdot x_{1}+P_{2}\cdot x_{2}+x_{5}=P}$ (augmented constraint) ${\displaystyle x_{1},x_{2},x_{3},x_{4},x_{5}\geq 0}$.
where ${\displaystyle x_{3},x_{4},x_{5}}$ are (non-negative) slack variables, representing in this example the unused area, the amount of unused fertilizer, and the amount of unused pesticide.
In matrix form this becomes:
Maximize ${\displaystyle z}$:
${\displaystyle {\begin{bmatrix}1&-S_{1}&-S_{2}&0&0&0\\0&1&1&1&0&0\\0&F_{1}&F_{2}&0&1&0\\0&P_{1}&P_{2}&0&0&1\\\end{bmatrix}}{\begin{bmatrix}z\\x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}}={\begin{bmatrix}0\\L\\F\\P\end{bmatrix}},\,{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}}\geq 0.}$
## Duality
Every linear programming problem, referred to as a primal problem, can be converted into a dual problem, which provides an upper bound to the optimal value of the primal problem. In matrix form, we can express the primal problem as:
Maximize cTx subject to Axb, x ≥ 0;
with the corresponding symmetric dual problem,
Minimize bTy subject to ATyc, y ≥ 0.
An alternative primal formulation is:
Maximize cTx subject to Axb;
with the corresponding asymmetric dual problem,
Minimize bTy subject to ATy = c, y ≥ 0.
There are two ideas fundamental to duality theory. One is the fact that (for the symmetric dual) the dual of a dual linear program is the original primal linear program. Additionally, every feasible solution for a linear program gives a bound on the optimal value of the objective function of its dual. The weak duality theorem states that the objective function value of the dual at any feasible solution is always greater than or equal to the objective function value of the primal at any feasible solution. The strong duality theorem states that if the primal has an optimal solution, x*, then the dual also has an optimal solution, y*, and cTx*=bTy*.
A linear program can also be unbounded or infeasible. Duality theory tells us that if the primal is unbounded then the dual is infeasible by the weak duality theorem. Likewise, if the dual is unbounded, then the primal must be infeasible. However, it is possible for both the dual and the primal to be infeasible. As an example, consider the linear program:
Maximize: ${\displaystyle 2x_{1}-x_{2}}$ Subject to: ${\displaystyle x_{1}-x_{2}\leq 1}$ ${\displaystyle -x_{1}+x_{2}\leq -2}$ ${\displaystyle x_{1},x_{2}\geq 0}$.
### Example
Revisit the above example of the farmer who may grow wheat and barley with the set provision of some L land, F fertilizer and P pesticide. Assume now that y unit prices for each of these means of production (inputs) are set by a planning board. The planning board's job is to minimize the total cost of procuring the set amounts of inputs while providing the farmer with a floor on the unit price of each of his crops (outputs), S1 for wheat and S2 for barley. This corresponds to the following linear programming problem:
Minimize: ${\displaystyle L\cdot y_{L}+F\cdot y_{F}+P\cdot y_{P}}$ (minimize the total cost of the means of production as the "objective function") subject to: ${\displaystyle y_{L}+F_{1}\cdot y_{F}+P_{1}\cdot y_{P}\geq S_{1}}$ (the farmer must receive no less than S1 for his wheat) ${\displaystyle y_{L}+F_{2}\cdot y_{F}+P_{2}\cdot y_{P}\geq S_{2}}$ (the farmer must receive no less than S2 for his barley) ${\displaystyle y_{L},y_{F},y_{P}\geq 0}$ (prices cannot be negative).
In matrix form this becomes:
Minimize: ${\displaystyle {\begin{bmatrix}L&F&P\end{bmatrix}}{\begin{bmatrix}y_{L}\\y_{F}\\y_{P}\end{bmatrix}}}$
subject to: ${\displaystyle {\begin{bmatrix}1&F_{1}&P_{1}\\1&F_{2}&P_{2}\end{bmatrix}}{\begin{bmatrix}y_{L}\\y_{F}\\y_{P}\end{bmatrix}}\geq {\begin{bmatrix}S_{1}\\S_{2}\end{bmatrix}},\,{\begin{bmatrix}y_{L}\\y_{F}\\y_{P}\end{bmatrix}}\geq 0.}$
The primal problem deals with physical quantities. With all inputs available in limited quantities, and assuming the unit prices of all outputs is known, what quantities of outputs to produce so as to maximize total revenue? The dual problem deals with economic values. With floor guarantees on all output unit prices, and assuming the available quantity of all inputs is known, what input unit pricing scheme to set so as to minimize total expenditure?
To each variable in the primal space corresponds an inequality to satisfy in the dual space, both indexed by output type. To each inequality to satisfy in the primal space corresponds a variable in the dual space, both indexed by input type.
The coefficients that bound the inequalities in the primal space are used to compute the objective in the dual space, input quantities in this example. The coefficients used to compute the objective in the primal space bound the inequalities in the dual space, output unit prices in this example.
Both the primal and the dual problems make use of the same matrix. In the primal space, this matrix expresses the consumption of physical quantities of inputs necessary to produce set quantities of outputs. In the dual space, it expresses the creation of the economic values associated with the outputs from set input unit prices.
Since each inequality can be replaced by an equality and a slack variable, this means each primal variable corresponds to a dual slack variable, and each dual variable corresponds to a primal slack variable. This relation allows us to speak about complementary slackness.
### Another example
Sometimes, one may find it more intuitive to obtain the dual program without looking at the program matrix. Consider the following linear program:
Minimize ${\displaystyle \sum _{i=1}^{m}{c_{i}x_{i}}+\sum _{j=1}^{n}{d_{j}t_{j}}}$ subject to ${\displaystyle \sum _{i=1}^{m}{a_{ij}x_{i}}+e_{j}t_{j}\geq g_{j}}$ , ${\displaystyle 1\leq j\leq n}$ ${\displaystyle f_{i}x_{i}+\sum _{j=1}^{n}{b_{ij}t_{j}}\geq h_{i}}$ , ${\displaystyle 1\leq i\leq m}$ ${\displaystyle x_{i}\geq 0,\,t_{j}\geq 0}$ , ${\displaystyle 1\leq i\leq m,1\leq j\leq n}$
We have m + n conditions and all variables are non-negative. We shall define m + n dual variables: yj and si. We get:
Minimize ${\displaystyle \sum _{i=1}^{m}{c_{i}x_{i}}+\sum _{j=1}^{n}{d_{j}t_{j}}}$ subject to ${\displaystyle \sum _{i=1}^{m}{a_{ij}x_{i}}\cdot y_{j}+e_{j}t_{j}\cdot y_{j}\geq g_{j}\cdot y_{j}}$ , ${\displaystyle 1\leq j\leq n}$ ${\displaystyle f_{i}x_{i}\cdot s_{i}+\sum _{j=1}^{n}{b_{ij}t_{j}}\cdot s_{i}\geq h_{i}\cdot s_{i}}$ , ${\displaystyle 1\leq i\leq m}$ ${\displaystyle x_{i}\geq 0,\,t_{j}\geq 0}$ , ${\displaystyle 1\leq i\leq m,1\leq j\leq n}$ ${\displaystyle y_{j}\geq 0,\,s_{i}\geq 0}$ , ${\displaystyle 1\leq j\leq n,1\leq i\leq m}$
Since this is a minimization problem, we would like to obtain a dual program that is a lower bound of the primal. In other words, we would like the sum of all right hand side of the constraints to be the maximal under the condition that for each primal variable the sum of its coefficients do not exceed its coefficient in the linear function. For example, x1 appears in n + 1 constraints. If we sum its constraints' coefficients we get a1,1y1 + a1,2y2 + ... + a1,nyn + f1s1. This sum must be at most c1. As a result, we get:
Maximize ${\displaystyle \sum _{j=1}^{n}{g_{j}y_{j}}+\sum _{i=1}^{m}{h_{i}s_{i}}}$ subject to ${\displaystyle \sum _{j=1}^{n}{a_{ij}y_{j}}+f_{i}s_{i}\leq c_{i}}$ , ${\displaystyle 1\leq i\leq m}$ ${\displaystyle e_{j}y_{j}+\sum _{i=1}^{m}{b_{ij}s_{i}}\leq d_{j}}$ , ${\displaystyle 1\leq j\leq n}$ ${\displaystyle y_{j}\geq 0,\,s_{i}\geq 0}$ , ${\displaystyle 1\leq j\leq n,1\leq i\leq m}$
Note that we assume in our calculations steps that the program is in standard form. However, any linear program may be transformed to standard form and it is therefore not a limiting factor.
## Variations
### Strict Inequality Constraints
One can allow for strict inequality constraints. In the general case, where only some inequalities are strict, we get a linear program of the form
Maximize: cTx,
subject to: A1 xb1, A2 x <b2,
Notice that equality constraints are nothing but a special case of non-strict inequality constraints.[5]
### Covering/packing dualities
A covering LP is a linear program of the form:
Minimize: bTy,
subject to: ATyc, y ≥ 0,
such that the matrix A and the vectors b and c are non-negative.
The dual of a covering LP is a packing LP, a linear program of the form:
Maximize: cTx,
subject to: Axb, x ≥ 0,
such that the matrix A and the vectors b and c are non-negative.
#### Examples
Covering and packing LPs commonly arise as a linear programming relaxation of a combinatorial problem and are important in the study of approximation algorithms.[6] For example, the LP relaxations of the set packing problem, the independent set problem, and the matching problem are packing LPs. The LP relaxations of the set cover problem, the vertex cover problem, and the dominating set problem are also covering LPs.
Finding a fractional coloring of a graph is another example of a covering LP. In this case, there is one constraint for each vertex of the graph and one variable for each independent set of the graph.
## Complementary slackness
It is possible to obtain an optimal solution to the dual when only an optimal solution to the primal is known using the complementary slackness theorem. The theorem states:
Suppose that x = (x1x2, ... , xn) is primal feasible and that y = (y1y2, ... , ym) is dual feasible. Let (w1w2, ..., wm) denote the corresponding primal slack variables, and let (z1z2, ... , zn) denote the corresponding dual slack variables. Then x and y are optimal for their respective problems if and only if
• xj zj = 0, for j = 1, 2, ... , n, and
• wi yi = 0, for i = 1, 2, ... , m.
So if the i-th slack variable of the primal is not zero, then the i-th variable of the dual is equal to zero. Likewise, if the j-th slack variable of the dual is not zero, then the j-th variable of the primal is equal to zero.
This necessary condition for optimality conveys a fairly simple economic principle. In standard form (when maximizing), if there is slack in a constrained primal resource (i.e., there are "leftovers"), then additional quantities of that resource must have no value. Likewise, if there is slack in the dual (shadow) price non-negativity constraint requirement, i.e., the price is not zero, then there must be scarce supplies (no "leftovers").
## Theory
### Existence of optimal solutions
Geometrically, the linear constraints define the feasible region, which is a convex polyhedron. A linear function is a convex function, which implies that every local minimum is a global minimum; similarly, a linear function is a concave function, which implies that every local maximum is a global maximum.
An optimal solution need not exist, for two reasons. First, if two constraints are inconsistent, then no feasible solution exists: For instance, the constraints x ≥ 2 and x ≤ 1 cannot be satisfied jointly; in this case, we say that the LP is infeasible. Second, when the polytope is unbounded in the direction of the gradient of the objective function (where the gradient of the objective function is the vector of the coefficients of the objective function), then no optimal value is attained.
### Optimal vertices (and rays) of polyhedra
Otherwise, if a feasible solution exists and if the (linear) objective function is bounded, then the optimum value is always attained on the boundary of optimal level-set, by the maximum principle for convex functions (alternatively, by the minimum principle for concave functions): Recall that linear functions are both convex and concave. However, some problems have distinct optimal solutions: For example, the problem of finding a feasible solution to a system of linear inequalities is a linear programming problem in which the objective function is the zero function (that is, the constant function taking the value zero everywhere): For this feasibility problem with the zero-function for its objective-function, if there are two distinct solutions, then every convex combination of the solutions is a solution.
The vertices of the polytope are also called basic feasible solutions. The reason for this choice of name is as follows. Let d denote the number of variables. Then the fundamental theorem of linear inequalities implies (for feasible problems) that for every vertex x* of the LP feasible region, there exists a set of d (or fewer) inequality constraints from the LP such that, when we treat those d constraints as equalities, the unique solution is x*. Thereby we can study these vertices by means of looking at certain subsets of the set of all constraints (a discrete set), rather than the continuum of LP solutions. This principle underlies the simplex algorithm for solving linear programs.
## Algorithms
In a linear programming problem, a series of linear constraints produces a convex feasible region of possible values for those variables. In the two-variable case this region is in the shape of a convex simple polygon.
### Basis exchange algorithms
#### Simplex algorithm of Dantzig
The simplex algorithm, developed by George Dantzig in 1947, solves LP problems by constructing a feasible solution at a vertex of the polytope and then walking along a path on the edges of the polytope to vertices with non-decreasing values of the objective function until an optimum is reached for sure. In many practical problems, "stalling" occurs: Many pivots are made with no increase in the objective function.[7][8] In rare practical problems, the usual versions of the simplex algorithm may actually "cycle".[8] To avoid cycles, researchers developed new pivoting rules.[9][10][7][8][11][12]
In practice, the simplex algorithm is quite efficient and can be guaranteed to find the global optimum if certain precautions against cycling are taken. The simplex algorithm has been proved to solve "random" problems efficiently, i.e. in a cubic number of steps,[13] which is similar to its behavior on practical problems.[7][14]
However, the simplex algorithm has poor worst-case behavior: Klee and Minty constructed a family of linear programming problems for which the simplex method takes a number of steps exponential in the problem size.[7][10][11] In fact, for some time it was not known whether the linear programming problem was solvable in polynomial time, i.e. of complexity class P.
#### Criss-cross algorithm
Like the simplex algorithm of Dantzig, the criss-cross algorithm is a basis-exchange algorithm that pivots between bases. However, the criss-cross algorithm need not maintain feasibility, but can pivot rather from a feasible basis to an infeasible basis. The criss-cross algorithm does not have polynomial time-complexity for linear programming. Both algorithms visit all 2D corners of a (perturbed) cube in dimension D, the Klee–Minty cube, in the worst case.[12][15]
### Interior point
In contrast to the simplex algorithm, which finds an optimal solution by traversing the edges between vertices on a polyhedral set, interior-point methods move through the interior of the feasible region.
#### Ellipsoid algorithm, following Khachiyan
This is the first worst-case polynomial-time algorithm for linear programming. To solve a problem which has n variables and can be encoded in L input bits, this algorithm uses O(n4L) pseudo-arithmetic operations on numbers with O(L) digits. Khachiyan's algorithm and his long standing issue was resolved by Leonid Khachiyan in 1979 with the introduction of the ellipsoid method. The convergence analysis has (real-number) predecessors, notably the iterative methods developed by Naum Z. Shor and the approximation algorithms by Arkadi Nemirovski and D. Yudin.
#### Projective algorithm of Karmarkar
Khachiyan's algorithm was of landmark importance for establishing the polynomial-time solvability of linear programs. The algorithm was not a computational break-through, as the simplex method is more efficient for all but specially constructed families of linear programs.
However, Khachiyan's algorithm inspired new lines of research in linear programming. In 1984, N. Karmarkar proposed a projective method for linear programming. Karmarkar's algorithm improved on Khachiyan's worst-case polynomial bound (giving ${\displaystyle O(n^{3.5}L)}$). Karmarkar claimed that his algorithm was much faster in practical LP than the simplex method, a claim that created great interest in interior-point methods.[16] Since Karmarkar's discovery, many interior-point methods have been proposed and analyzed.
#### Affine scaling
Affine scaling is one of the oldest interior point methods to be developed. It was developed in the Soviet Union in the mid-1960s, but didn't receive much attention until the discovery of Karmarkar's algorithm, after which affine scaling was reinvented multiple times and presented as a simplified version of Karmarkar's. Affine scaling amounts to doing gradient descent steps within the feasible region, while rescaling the problem to make sure the steps move toward the optimum faster.[17]
#### Path-following algorithms
For both theoretical and practical purposes, barrier function or path-following methods have been the most popular interior point methods since the 1990s.[18]
### Comparison of interior-point methods versus simplex algorithms
The current opinion is that the efficiency of good implementations of simplex-based methods and interior point methods are similar for routine applications of linear programming.[18] However, for specific types of LP problems, it may be that one type of solver is better than another (sometimes much better), and that the structure of the solutions generated by interior point methods versus simplex-based methods are significantly different with the support set of active variables being typically smaller for the later one.[19]
LP solvers are in widespread use for optimization of various problems in industry, such as optimization of flow in transportation networks.[20]
### Approximate algorithms for covering/packing LPs
Covering and packing LPs can be solved approximately in nearly-linear time. That is, if matrix A is of dimension n×m and has N non-zero entries, then there exist algorithms that run in time O(N·(log N)O(1)/εO(1)) and produce O(1±ε) approximate solutions to given covering and packing LPs. The best known sequential algorithm of this kind runs in time O(N + (log N)·(n+m)/ε2),[21] and the best known parallel algorithm of this kind runs in O((log N)2/ε3) iterations, each requiring only a matrix-vector multiplication which is highly parallelizable.[22]
## Open problems and recent work
Unsolved problem in computer science: Does linear programming admit a strongly polynomial-time algorithm? (more unsolved problems in computer science)
There are several open problems in the theory of linear programming, the solution of which would represent fundamental breakthroughs in mathematics and potentially major advances in our ability to solve large-scale linear programs.
• Does LP admit a strongly polynomial-time algorithm?
• Does LP admit a strongly polynomial algorithm to find a strictly complementary solution?
• Does LP admit a polynomial algorithm in the real number (unit cost) model of computation?
This closely related set of problems has been cited by Stephen Smale as among the 18 greatest unsolved problems of the 21st century. In Smale's words, the third version of the problem "is the main unsolved problem of linear programming theory." While algorithms exist to solve linear programming in weakly polynomial time, such as the ellipsoid methods and interior-point techniques, no algorithms have yet been found that allow strongly polynomial-time performance in the number of constraints and the number of variables. The development of such algorithms would be of great theoretical interest, and perhaps allow practical gains in solving large LPs as well.
Although the Hirsch conjecture was recently disproved for higher dimensions, it still leaves the following questions open.
• Are there pivot rules which lead to polynomial-time Simplex variants?
• Do all polytopal graphs have polynomially bounded diameter?
These questions relate to the performance analysis and development of Simplex-like methods. The immense efficiency of the Simplex algorithm in practice despite its exponential-time theoretical performance hints that there may be variations of Simplex that run in polynomial or even strongly polynomial time. It would be of great practical and theoretical significance to know whether any such variants exist, particularly as an approach to deciding if LP can be solved in strongly polynomial time.
The Simplex algorithm and its variants fall in the family of edge-following algorithms, so named because they solve linear programming problems by moving from vertex to vertex along edges of a polytope. This means that their theoretical performance is limited by the maximum number of edges between any two vertices on the LP polytope. As a result, we are interested in knowing the maximum graph-theoretical diameter of polytopal graphs. It has been proved that all polytopes have subexponential diameter. The recent disproof of the Hirsch conjecture is the first step to prove whether any polytope has superpolynomial diameter. If any such polytopes exist, then no edge-following variant can run in polynomial time. Questions about polytope diameter are of independent mathematical interest.
Simplex pivot methods preserve primal (or dual) feasibility. On the other hand, criss-cross pivot methods do not preserve (primal or dual) feasibility—they may visit primal feasible, dual feasible or primal-and-dual infeasible bases in any order. Pivot methods of this type have been studied since the 1970s. Essentially, these methods attempt to find the shortest pivot path on the arrangement polytope under the linear programming problem. In contrast to polytopal graphs, graphs of arrangement polytopes are known to have small diameter, allowing the possibility of strongly polynomial-time criss-cross pivot algorithm without resolving questions about the diameter of general polytopes.[12]
## Integer unknowns
If all of the unknown variables are required to be integers, then the problem is called an integer programming (IP) or integer linear programming (ILP) problem. In contrast to linear programming, which can be solved efficiently in the worst case, integer programming problems are in many practical situations (those with bounded variables) NP-hard. 0-1 integer programming or binary integer programming (BIP) is the special case of integer programming where variables are required to be 0 or 1 (rather than arbitrary integers). This problem is also classified as NP-hard, and in fact the decision version was one of Karp's 21 NP-complete problems.
If only some of the unknown variables are required to be integers, then the problem is called a mixed integer programming (MIP) problem. These are generally also NP-hard because they are even more general than ILP programs.
There are however some important subclasses of IP and MIP problems that are efficiently solvable, most notably problems where the constraint matrix is totally unimodular and the right-hand sides of the constraints are integers or – more general – where the system has the total dual integrality (TDI) property.
Advanced algorithms for solving integer linear programs include:
Such integer-programming algorithms are discussed by Padberg and in Beasley.
## Integral linear programs
A linear program in real variables is said to be integral if it has at least one optimal solution which is integral. Likewise, a polyhedron ${\displaystyle P=\{x\mid Ax\geq 0\}}$ is said to be integral if for all bounded feasible objective functions c, the linear program ${\displaystyle \{\max cx\mid x\in P\}}$ has an optimum ${\displaystyle x^{*}}$ with integer coordinates. As observed by Edmonds and Giles in 1977, one can equivalently say that the polyhedron ${\displaystyle P}$ is integral if for every bounded feasible integral objective function c, the optimal value of the linear program ${\displaystyle \{\max cx\mid x\in P\}}$ is an integer.
Integral linear programs are of central importance in the polyhedral aspect of combinatorial optimization since they provide an alternate characterization of a problem. Specifically, for any problem, the convex hull of the solutions is an integral polyhedron; if this polyhedron has a nice/compact description, then we can efficiently find the optimal feasible solution under any linear objective. Conversely, if we can prove that a linear programming relaxation is integral, then it is the desired description of the convex hull of feasible (integral) solutions.
Note that terminology is not consistent throughout the literature, so one should be careful to distinguish the following two concepts,
• in an integer linear program, described in the previous section, variables are forcibly constrained to be integers, and this problem is NP-hard in general,
• in an integral linear program, described in this section, variables are not constrained to be integers but rather one has proven somehow that the continuous problem always has an integral optimal value (assuming c is integral), and this optimal value may be found efficiently since all polynomial-size linear programs can be solved in polynomial time.
One common way of proving that a polyhedron is integral is to show that it is totally unimodular. There are other general methods including the integer decomposition property and total dual integrality. Other specific well-known integral LPs include the matching polytope, lattice polyhedra, submodular flow polyhedra, and the intersection of 2 generalized polymatroids/g-polymatroids --- e.g. see Schrijver 2003.
A bounded integral polyhedron is sometimes called a convex lattice polytope, particularly in two dimensions.
## Solvers and scripting (programming) languages
Pyomo BSD An open-source modeling language for large-scale linear, mixed integer and nonlinear optimization
Cassowary constraint solver LGPL an incremental constraint solving toolkit that efficiently solves systems of linear equalities and inequalities
CLP CPL an LP solver from COIN-OR
glpk GPL GNU Linear Programming Kit, an LP/MILP solver with a native C API and numerous (15) third-party wrappers for other languages. Specialist support for flow networks. Bundles the AMPL-like GNU MathProg modelling language and translator.
Qoca GPL a library for incrementally solving systems of linear equations with various goal functions
QSopt-Exact GPL a C-based solver library using GNU MP. [23]
R-Project GPL a programming language and software environment for statistical computing and graphics
MINTO (Mixed Integer Optimizer, an integer programming solver which uses branch and bound algorithm) has publicly available source code[24] but is not open source.
Proprietary:
Name Brief info
AIMMS
AMPL A popular modeling language for large-scale linear, mixed integer and nonlinear optimisation with a free student limited version available (500 variables and 500 constraints).
APMonitor API to MATLAB and Python. Solve example Linear Programming (LP) problems through MATLAB, Python, or a web-interface.
CPLEX Popular solver with an API for several programming languages, and also has a modelling language and works with AIMMS, AMPL, GAMS, MPL, OpenOpt, OPL Development Studio, and TOMLAB. Free for academic use.
Excel Solver Function A nonlinear solver adjusted to spreadsheets in which function evaluations are based on the recalculating cells. Basic version available as a standard add-on for Excel.
FortMP
GAMS
Gurobi Solver with parallel algorithms for large-scale linear programs, quadratic programs and mixed-integer programs. Free for academic use.
IMSL Numerical Libraries Collections of math and statistical algorithms available in C/C++, Fortran, Java and C#/.NET. Optimization routines in the IMSL Libraries include unconstrained, linearly and nonlinearly constrained minimizations, and linear programming algorithms.
LINDO Solver with an API for large scale optimization of linear, integer, quadratic, conic and general nonlinear programs with stochastic programming extensions. It offers a global optimization procedure for finding guaranteed globally optimal solution to general nonlinear programs with continuous and discrete variables. It also has a statistical sampling API to integrate Monte-Carlo simulations into an optimization framework. It has an algebraic modeling language (LINGO) and allows modeling within a spreadsheet (What'sBest).
Maple A general-purpose programming-language for symbolic and numerical computing.
MATLAB A general-purpose and matrix-oriented programming-language for numerical computing. Linear programming in MATLAB requires the Optimization Toolbox in addition to the base MATLAB product; available routines include INTLINPROG and LINPROG
Mathcad A WYSIWYG math editor. It has functions for solving both linear and nonlinear optimization problems.
Mathematica A general-purpose programming-language for mathematics, including symbolic and numerical capabilities.
MOSEK A solver for large scale optimization with API for several languages (C++,java,.net, Matlab and python).
NAG Numerical Library A collection of mathematical and statistical routines developed by the Numerical Algorithms Group for multiple programming languages (C, C++, Fortran, Visual Basic, Java and C#) and packages (MATLAB, Excel, R, LabVIEW). The Optimization chapter of the NAG Library includes routines for linear programming problems with both sparse and non-sparse linear constraint matrices, together with routines for the optimization of quadratic, nonlinear, sums of squares of linear or nonlinear functions with nonlinear, bounded or no constraints. The NAG Library has routines for both local and global optimization, and for continuous or integer problems.
NMath Stats A general-purpose .NET statistical library containing a simplex solver.[25]
QSopt A C-based floating-point solver offered as a callable library and command-line and UI executables. Binaries available for Windows, Mac OS and several UNIX flavors. Free for research purposes. Compatible with MPS and LP indistry-standard formats.[26]
OptimJ A Java-based modeling language for optimization with a free version available.[27][28]
SAS/OR A suite of solvers for Linear, Integer, Nonlinear, Derivative-Free, Network, Combinatorial and Constraint Optimization; the Algebraic modeling language OPTMODEL; and a variety of vertical solutions aimed at specific problems/markets, all of which are fully integrated with the SAS System.
SCIP A general-purpose constraint integer programming solver with an emphasis on MIP. Compatible with Zimpl modelling language. Free for academic use and available in source code.
XPRESS Solver for large-scale linear programs, quadratic programs, general nonlinear and mixed-integer programs. Has API for several programming languages, also has a modelling language Mosel and works with AMPL, GAMS. Free for academic use.
VisSim A visual block diagram language for simulation of dynamical systems.
## Notes
1. ^ a b Gerard Sierksma (2001). Linear and Integer Programming: Theory and Practice, Second Edition. CRC Press. p. 1. ISBN 978-0-8247-0673-9.
2. ^ a b Alexander Schrijver (1998). Theory of Linear and Integer Programming. John Wiley & Sons. pp. 221–222. ISBN 978-0-471-98232-6.
3. ^ "Reminiscences about the origins of linear programming" (PDF). Operations Research Letter. April 1982. doi:10.1016/0167-6377(82)90043-8.
4. ^ "Reminiscences about the origins of linear programming" (PDF). Operations Research Letter. 1 (2): 43–48. April 1982. doi:10.1016/0167-6377(82)90043-8.
5. ^
6. ^ Vazirani (2001, p. 112)
7. ^ a b c d Dantzig & Thapa (2003)
8. ^ a b c Padberg (1999)
9. ^ Bland (1977)
10. ^ a b Murty (1983)
11. ^ a b Papadimitriou & Steiglitz
12. ^ a b c Fukuda & Terlaky (1997): Fukuda, Komei; Terlaky, Tamás (1997). Thomas M. Liebling and Dominique de Werra, eds. "Criss-cross methods: A fresh view on pivot algorithms". Mathematical Programming: Series B. Amsterdam: North-Holland Publishing Co. 79 (1—3): 369–395. MR 1464775. doi:10.1007/BF02614325.
13. ^ Borgwardt (1987)
14. ^ Todd (2002)
15. ^ Roos (1990): Roos, C. (1990). "An exponential example for Terlaky's pivoting rule for the criss-cross simplex method". Mathematical Programming. Series A. 46 (1): 79–84. MR 1045573. doi:10.1007/BF01585729.
16. ^ Strang, Gilbert (1 June 1987). "Karmarkar's algorithm and its place in applied mathematics". The Mathematical Intelligencer. New York: Springer. 9 (2): 4–10. ISSN 0343-6993. MR '''883185'''. doi:10.1007/BF03025891.
17. ^ Vanderbei (2001), pp. 333–347
18. ^ a b Gondzio & Terlaky (1996)
19. ^ Illés, Tibor; Terlaky, Tamás (2002). "Pivot versus interior point methods: Pros and cons". European Journal of Operational Research. 140 (2): 170. doi:10.1016/S0377-2217(02)00061-9.
20. ^ For solving network-flow problems in transportation networks, specialized implementations of the simplex algorithm can dramatically improve its efficiency. Dantzig & Thapa (2003)
21. ^ Christos Koufogiannakis; Neal E. Young (2013). "A Nearly Linear-Time PTAS for Explicit Fractional Packing and Covering Linear Programs". Algorithmica. 70: 648–674. arXiv:. doi:10.1007/s00453-013-9771-6.
22. ^ Zeyuan Allen-Zhu; Lorenzo Orecchia (2015). Using Optimization to Break the Epsilon Barrier: A Faster and Simpler Width-Independent Algorithm for Solving Positive Linear Programs in Parallel. ACM-SIAM Symposium on Discrete Algorithms. arXiv:.
24. ^
25. ^ "C# Linear Programming". centerspace.net.
27. ^ http://www.in-ter-trans.eu/resources/Zesch_Hellingrath_2010_Integrated+Production-Distribution+Planning.pdf OptimJ used in an optimization model for mixed-model assembly lines, University of Münster
28. ^ http://www.aaai.org/ocs/index.php/AAAI/AAAI10/paper/viewFile/1769/2076 OptimJ used in an Approximate Subgame-Perfect Equilibrium Computation Technique for Repeated Games
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• Gondzio, Jacek; Terlaky, Tamás (1996). "3 A computational view of interior point methods". In J. E. Beasley. Advances in linear and integer programming. Oxford Lecture Series in Mathematics and its Applications. 4. New York: Oxford University Press. pp. 103–144. MR 1438311. Postscript file at website of Gondzio and at McMaster University website of Terlaky.
• Murty, Katta G. (1983). Linear programming. New York: John Wiley & Sons, Inc. pp. xix+482. ISBN 0-471-09725-X. MR 720547. (comprehensive reference to classical approaches).
• Evar D. Nering and Albert W. Tucker, 1993, Linear Programs and Related Problems, Academic Press. (elementary)
• M. Padberg, Linear Optimization and Extensions, Second Edition, Springer-Verlag, 1999. (carefully written account of primal and dual simplex algorithms and projective algorithms, with an introduction to integer linear programming --- featuring the traveling salesman problem for Odysseus.)
• Christos H. Papadimitriou and Kenneth Steiglitz, Combinatorial Optimization: Algorithms and Complexity, Corrected republication with a new preface, Dover. (computer science)
• Michael J. Todd (February 2002). "The many facets of linear programming". Mathematical Programming. 91 (3): 417–436. doi:10.1007/s101070100261. (Invited survey, from the International Symposium on Mathematical Programming.)
• Vanderbei, Robert J. (2001). Linear Programming: Foundations and Extensions. Springer Verlag.
• Vazirani, Vijay V. (2001). Approximation Algorithms. Springer-Verlag. ISBN 3-540-65367-8. (Computer science)
A reader may consider beginning with Nering and Tucker, with the first volume of Dantzig and Thapa, or with Williams.
• Dmitris Alevras and Manfred W. Padberg, Linear Optimization and Extensions: Problems and Solutions, Universitext, Springer-Verlag, 2001. (Problems from Padberg with solutions.)
• Mark de Berg, Marc van Kreveld, Mark Overmars, and Otfried Schwarzkopf (2000). Computational Geometry (2nd revised ed.). Springer-Verlag. ISBN 3-540-65620-0. Chapter 4: Linear Programming: pp. 63–94. Describes a randomized half-plane intersection algorithm for linear programming.
• Michael R. Garey and David S. Johnson (1979). Computers and Intractability: A Guide to the Theory of NP-Completeness. W.H. Freeman. ISBN 0-7167-1045-5. A6: MP1: INTEGER PROGRAMMING, pg.245. (computer science, complexity theory)
• Bernd Gärtner, Jiří Matoušek (2006). Understanding and Using Linear Programming, Berlin: Springer. ISBN 3-540-30697-8 (elementary introduction for mathematicians and computer scientists)
• Cornelis Roos, Tamás Terlaky, Jean-Philippe Vial, Interior Point Methods for Linear Optimization, Second Edition, Springer-Verlag, 2006. (Graduate level)
• Alexander Schrijver (2003). Combinatorial optimization: polyhedra and efficiency. Springer.
• Alexander Schrijver, Theory of Linear and Integer Programming. John Wiley & sons, 1998, ISBN 0-471-98232-6 (mathematical)
• H. P. Williams, Model Building in Mathematical Programming, Third revised Edition, 1990. (Modeling)
• Stephen J. Wright, 1997, Primal-Dual Interior-Point Methods, SIAM. (Graduate level)
• Yinyu Ye, 1997, Interior Point Algorithms: Theory and Analysis, Wiley. (Advanced graduate-level)
• Ziegler, Günter M., Chapters 1–3 and 6–7 in Lectures on Polytopes, Springer-Verlag, New York, 1994. (Geometry)
| 12,147 | 50,051 |
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Problem on projection of gravitational force down an inclined plane
A block rests on an inclined plane of angle $\theta$, as shown. Gravity provides a force $\bfF = (0, -m g)$ on the block.
(a) What is the component of $\bfF$ in the downhill direction?
(b) What is the projection of $\bfF$ in the same direction?
• Solution
We begin by drawing the relevant vectors.
We note that $\bfF = (0, -m g)$. A vector pointing downhill as shown is $\bfd = (\cos \theta, - \sin \theta)$.
Recall that
We now compute that
Hence, $$\text{comp}_\bfd \bfF = mg \sin \theta.$$
Recall that
Having already computed $\bfF \cdot \bfd$ and $|\bfd|$, we can write that $$\text{proj}_{\bfd}\ \bfF = mg \sin \theta (\cos \theta, - \sin \theta).$$
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https://forum.piedao.org/answers/13450127-one-thousand-tonnes-1000-t-one-t-equals-10-cubed-kg-of-sand-contains-about-a-trillion-10-super-12
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# One thousand tonnes (1000 t, one t equals 10 cubed kg) of sand contains about a trillion (10 super 12) grains of sand. How many tonnes of sand are needed to provide 1 mol of grains of sand? (b) Assuming the volume of a grain of sand is 1.0 mm3 and the land area of the continental United States is 3.6 multiplication 10 super six square miles, how deep would the sand pile over the United States be if this area were evenly covered with 1.0 mol of grains of sand?
Since 1 mole of sand will contain Avagadro's Number of sand particles (by definition of 1 mole)
Thus we have
1 mole of sand = sand particles
Thus in number of trillion sand particles we have no of trillion sand particles in 12 mole is
Now since it is given that mass of 1 trillion sand particles is 1000 tonnes Thus the mass of trillion sand particles is
Part 2)
Since it is given that volume of 1 sand particle is thus the volume of 1 mole of sand is volume of sand particles
Thus volume of 1 mole is
Now since the Area of united states is
Thus the depth of the sand pile is
## Related Questions
What is the result of adding these two equations?\begin{aligned} -5x-9y &= 3 \\\\ 5x-9y &= -2 \end{aligned}
−5x−9y
5x−9y
=3
=−2
-18y=1
Step-by-step explanation:
What is the upper quartile of 36 ,73,28,36,80,50
Put the numbers in order
28, 36, 36, 50, 73, 80
|. |. |
The upper quartile is 73
the upper quartile is 80
Find the zeros of the function
k(x) = -5x² - 125
The zeros of k are x= □ and x= □
x = ±5i
General Formulas and Concepts:
Pre-Alg
• Order of Operations: BPEMDAS
Alg II
• √-1 is imaginary number i
Step-by-step explanation:
Step 1: Define function
k(x) = -5x² - 125
Step 2: Find roots
1. Set function equal to 0: 0 = -5x² - 125
2. Factor out -5: 0 = -5(x² + 25)
3. Divide both sides by -5: 0 = x² + 25
4. Subtract 25 on both sides: -25 = x²
5. Rewrite: x² = -25
6. Square root both sides: x = ±√-25
7. Rewrite: x = √-1 · ±√25
8. Evaluate: x = ±5i
Billy wants to buy 657 mugs. The mugs are sold for a price of 3 for $4. How much must Billy pay for all the mugs? ### Answers 219 *$4 = \$876
875 is the answer to your question so have a nice day
Emily drove to town with an average speed of 32 miles per hour, and then back home with an average speed of 38 miles per hour. If her total traveling time was 42 minutes, how far is it from home to town?
• 12.16 miles
Step-by-step explanation:
Speed - s, Distance - d, Time - t
Equation of time is:
• t = d/s
Given,
• s1 = 32 m/h, s2 = 38 m/h, t1 + t2 = 42 min = 42/60 h = 7/10 h
Total time is the sum of time values to and from the town:
• d/32 + d/38 = 7/10
• d/16 + d/19 = 7/5
• d(1/16 + 1/19) = 7/5
• d(16 + 19) = 7(16*19)/5
• 35d = 425.6
• d = 425.6/35
• d = 12.16
It is given that,
→ s1= 32 miles/h
→ s2 = 38 miles/h
Now t1 + t2 is,
→ 42 min
→ 42/60 hours
→ 7/10 hours
The formula we use,
→ Time = Distance/Speed
→ t = d/s
Then the total time is the,
Sum of time values to and from the town.
→ d/32 + d/38 = 7/10
→ d/16 + d/19 = 7/5
→ d{(1/16) + (1/19)} = 7/5
→ d(16 + 19) = (7/5) × (16 × 19)
→ 35d = 425.6
→ d = 425.6/35
→ d = 12.16
Hence, the distance is 12.16 miles.
The expression 2x(5x - 1) +(3x2 - 7x - 2) is simplified into the form ax 2+ bx + c.What are the values of the coefficients a,b,c?
a:
b:
C:
Step-by-step explanation:
2x (5x - 1) + (3x^2 - 7x - 2)
= 10x^2 - 2x + 3x^2 - 7x - 2
= 13x^2 - 9x - 2
a = 13
b = -9
c = 2
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Maths > Maths skills for age 3–4
# Maths skills for age 3–4
### Beginning to explore numbers
It’s never too early to explore numbers with your child. Real life often provides the best opportunities to develop early maths skills like counting and recognising numbers. These ideas will help your child to explore numbers, giving them a great start to developing early maths skills and concepts.
## Things to try with your child
#### 1. Listen to and sing songs and rhymes
Sing – even if it isn’t your strong point! Sing counting songs, such as '10 Green Bottles', '1, 2, Buckle My Shoe’ and ‘1, 2, 3, 4, 5, Once I Caught a Fish Alive’. Singing songs is a good way for children to become familiar with counting. Don't worry if they choose the same songs again and again!
Want a number rhyme to try? Download our ‘Three little frogs’ activity sheet >
#### 2. Talk about numbers around you
Numbers are all around us, from calendars to the remote control, the telephone to car registration plates. Try pointing out numbers when you’re out and about – on front doors, signs, the front of buses and train platforms. Talking about numbers around you early and often will show your child that numbers are part of everyday life. Choose a 'Number of the Week' and see how many times you can spot this number, around the house, out in the street or in the supermarket.
Share a book with your child. There are many fantastic books based around numbers, but any book can be used to help children develop early counting and number recognition skills. Take time to talk about what your child can see on each page. Count objects on the page, and compare the number of objects from page to page. Look at the page numbers and say them together.
Find the maths on every page with these free eBooks >
#### 4. Count as much as you can
Count whenever you can – count together, as well as letting your child see and hear you counting. Practise chanting numbers and, as your child’s confidence grows, start from different numbers – 5, 6, 7, etc. Count real objects – your child’s toy cars, pencils, shoes, or the number of stairs in your house. Don’t worry if your child remembers the answer – they can count to check!
Try to stick to a single type of object for each counting activity, and encourage your child to touch or pick up each object as they count it. Ask your child to help you sort cutlery or laundry, counting as you sort. When you go out for a walk, count your footsteps, the number of cars or houses you see etc.
#### 5. Get your hands dirty
Help your child to learn the numerals by exploring their shapes. Have fun forming numbers in sand with a stick, on the pavement with chalk or on sheets of paper with finger paints. Make numbers out of modelling clay. Form numbers out of small objects such as pieces of pasta or beads. Try holding your child’s finger and forming the number in the air. All these activities can help your child to become more familiar with numerals and enjoy themselves in the process!
#### 6. Play maths games
Try these fun games with your child to practise early maths skills and help to build your child's confidence. Most children love playing games and it's an easy way to support their learning.
### Match the Shape
Look at the picture and find all the hidden shapes.
### Whose is it?
Help each of the three bears find their things. Match each bear to their objects.
### Matching Pairs
Link all the muddled-up shoes to make pairs.
### Jack and the Beanstalk
Drag the pictures to put this story in order. Use the words 1st, 2nd, 3rd, and so on.
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# Trains Time | Multiple Choice Questions (MCQ) Choose PartPart A
Q1 - A bus is moving with a speed of 72 kmh, convert its speed into ms:
A. 10ms
B. 720ms
C. 30ms
D. 20ms
Detail - No Details
Q2 - The ratio of the speeds of two trains is 3:4. If the second train runs 300 km in 3 hours then what would be the speed of the first train?
A. 100 km/hr
B. 50 km/hr
C. 70 km/hr
D. 75 km/hr
Detail - No Details
Q3 - Two cars travel from city A to city B at a speed of 30 and 44 km/hr respectively. If one car takes 3.5 hours lesser time than the other car for the journey, then the distance between City A and City B is
A. 330 km
B. 396 km
C. 495 km
D. 264 km
Detail - No Details
Q4 - A car can cover a distance of 522 km on 36 liters of petrol. How far can it travel on 14 liters of petrol?
A. 213 km
B. 223 km
C. 203 km
D. 302 km
Detail - No Details
Q5 - A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km?
A. 220 km
B. 224 km
C. 230 km
D. 234 km
Detail - No Details
Q6 - In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is?
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours
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### How to Calculate the Circumference of a Circle
Adapted Mind Math is a unique online game platform that offers students the opportunity to learn mathematics in a customized learning environment. Making use of a challenge-based format, Adapted Mind Math pushes students to excel while offering a range of additional tutorials that help players grasp key concepts in algebra and geometry. Students will often find themselves working with circles when studying geometry, so here is how to calculate the circumference of a circle.
The circumference is the enclosing boundary of a circle. In essence, it is the continuous line that makes up the shape. To work out its length, you will need the values for the circle’s diameter and π.
The diameter is the length of a line passing from one side of the circle to the other through the circle’s center point. If the circle is drawn to scale, you can measure this manually. However, if it isn’t drawn to scale, it is likely that you will be provided with the diameter's measurement or that of the radius. If you have the radius, simply multiply it by two to find the diameter.
The number π is a numerical value for which the decimals run to infinity. Many calculators will provide you with a version of π to use in equations. However, if you don’t have a calculator with this feature, use the number 3.14. This is π to two decimal places.
Multiply the diameter and π together to find the circumference, ensuring your answer uses the same measurements as those given for the diameter.
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# 50439 (number)
50,439 (fifty thousand four hundred thirty-nine) is an odd five-digits composite number following 50438 and preceding 50440. In scientific notation, it is written as 5.0439 × 104. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 29,568 positive integers (up to 50439) that are relatively prime to 50439.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 21
• Digital Root 3
## Name
Short name 50 thousand 439 fifty thousand four hundred thirty-nine
## Notation
Scientific notation 5.0439 × 104 50.439 × 103
## Prime Factorization of 50439
Prime Factorization 3 × 17 × 23 × 43
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 50439 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 50,439 is 3 × 17 × 23 × 43. Since it has a total of 4 prime factors, 50,439 is a composite number.
## Divisors of 50439
1, 3, 17, 23, 43, 51, 69, 129, 391, 731, 989, 1173, 2193, 2967, 16813, 50439
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 76032 Sum of all the positive divisors of n s(n) 25593 Sum of the proper positive divisors of n A(n) 4752 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 224.586 Returns the nth root of the product of n divisors H(n) 10.6143 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 50,439 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 50,439) is 76,032, the average is 4,752.
## Other Arithmetic Functions (n = 50439)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 29568 Total number of positive integers not greater than n that are coprime to n λ(n) 3696 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5171 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 29,568 positive integers (less than 50,439) that are coprime with 50,439. And there are approximately 5,171 prime numbers less than or equal to 50,439.
## Divisibility of 50439
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 4 7 3
The number 50,439 is divisible by 3.
## Classification of 50439
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (50439)
Base System Value
2 Binary 1100010100000111
3 Ternary 2120012010
4 Quaternary 30110013
5 Quinary 3103224
6 Senary 1025303
8 Octal 142407
10 Decimal 50439
12 Duodecimal 25233
16 Hexadecimal c507
20 Vigesimal 661j
36 Base36 12x3
## Basic calculations (n = 50439)
### Multiplication
n×y
n×2 100878 151317 201756 252195
### Division
n÷y
n÷2 25219.5 16813 12609.8 10087.8
### Exponentiation
ny
n2 2544092721 128321492754519 6472407773045183841 326461775664626027756199
### Nth Root
y√n
2√n 224.586 36.9478 14.9862 8.72074
## 50439 as geometric shapes
### Circle
Radius = n
Diameter 100878 316918 7.9925e+09
### Sphere
Radius = n
Volume 5.37512e+14 3.197e+10 316918
### Square
Length = n
Perimeter 201756 2.54409e+09 71331.5
### Cube
Length = n
Surface area 1.52646e+10 1.28321e+14 87362.9
### Equilateral Triangle
Length = n
Perimeter 151317 1.10162e+09 43681.5
### Triangular Pyramid
Length = n
Surface area 4.4065e+09 1.51228e+13 41183.3
## Cryptographic Hash Functions
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10 January, 11:28
# An 114 kg man climbs up a 1 m high flight of stairs. The acceleration of gravity is 9.81 m/s 2. What is the increase in the gravitational potential energy of the man-earth system? Answer in units of kJ.
+1
1. 10 January, 13:20
0
1.118KJ
Explanation:
Potential energy is the energy possessed by a body by virtue of its position.
If the body has a mass (m), and climbs through height (h), the potential energy will be mass*acceleration due to gravity*height
P. E = mgh
Given m = 114kg, g = 9.81m/s² h = 1m
P. E = 114*9.81*1
P. E = 1,118.34Joules
P. E = 1.118KJ
This shows that increase in the gravitational potential energy of the man-earth system is 1.118KJ
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| 3.984375 | 4 |
CC-MAIN-2021-21
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| 0.899224 |
https://metanumbers.com/38210
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# 38210 (number)
38,210 (thirty-eight thousand two hundred ten) is an even five-digits composite number following 38209 and preceding 38211. In scientific notation, it is written as 3.821 × 104. The sum of its digits is 14. It has a total of 3 prime factors and 8 positive divisors. There are 15,280 positive integers (up to 38210) that are relatively prime to 38210.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 14
• Digital Root 5
## Name
Short name 38 thousand 210 thirty-eight thousand two hundred ten
## Notation
Scientific notation 3.821 × 104 38.21 × 103
## Prime Factorization of 38210
Prime Factorization 2 × 5 × 3821
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 38210 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 38,210 is 2 × 5 × 3821. Since it has a total of 3 prime factors, 38,210 is a composite number.
## Divisors of 38210
1, 2, 5, 10, 3821, 7642, 19105, 38210
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 68796 Sum of all the positive divisors of n s(n) 30586 Sum of the proper positive divisors of n A(n) 8599.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 195.474 Returns the nth root of the product of n divisors H(n) 4.44328 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 38,210 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 38,210) is 68,796, the average is 859,9.5.
## Other Arithmetic Functions (n = 38210)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 15280 Total number of positive integers not greater than n that are coprime to n λ(n) 3820 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4032 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 15,280 positive integers (less than 38,210) that are coprime with 38,210. And there are approximately 4,032 prime numbers less than or equal to 38,210.
## Divisibility of 38210
m n mod m 2 3 4 5 6 7 8 9 0 2 2 0 2 4 2 5
The number 38,210 is divisible by 2 and 5.
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (38210)
Base System Value
2 Binary 1001010101000010
3 Ternary 1221102012
4 Quaternary 21111002
5 Quinary 2210320
6 Senary 452522
8 Octal 112502
10 Decimal 38210
12 Duodecimal 1a142
20 Vigesimal 4faa
36 Base36 the
## Basic calculations (n = 38210)
### Multiplication
n×y
n×2 76420 114630 152840 191050
### Division
n÷y
n÷2 19105 12736.7 9552.5 7642
### Exponentiation
ny
n2 1460004100 55786756661000 2131611972016810000 81448893450762310100000
### Nth Root
y√n
2√n 195.474 33.6816 13.9812 8.24965
## 38210 as geometric shapes
### Circle
Diameter 76420 240081 4.58674e+09
### Sphere
Volume 2.33679e+14 1.8347e+10 240081
### Square
Length = n
Perimeter 152840 1.46e+09 54037.1
### Cube
Length = n
Surface area 8.76002e+09 5.57868e+13 66181.7
### Equilateral Triangle
Length = n
Perimeter 114630 6.322e+08 33090.8
### Triangular Pyramid
Length = n
Surface area 2.5288e+09 6.57453e+12 31198.3
## Cryptographic Hash Functions
md5 43bb4564e4258d621110838c7328c921 9d16f7d0fe8d3afb47801535d61c2f08a8b95c91 cd6643966dc81728b3542b59cadf6e723ce990f065c8dbf783dfab765832ec4b 3f0b7c8dfce10f4de5bb746e96bd3b84257bf0ce5233c6a1dd05ef7b62eceeecc00ea9424f06468705cc86495b5b2e4c7880dd9e10ab37e4cba125e0eb596063 16a2ee1f334675c6028f6675f619fcb02170f39c
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# math
posted by .
the legs of an isosceles triangle have lengths x+1 and -x+7. the base has length 3x-3. what is the length of the base?
a. 6
b. cannot be determined
c. 4
d. 3
• math -
we know that
x+1 = -x+7
x=3
so the base is 6
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## Tuesday, May 09, 2006
### Odd and even numbers - answers for VW
VW asked some questions about odd and even numbers in her comment to this post and since I live to torture you all, here are the questions and the answers. *grin*
1. Why does adding two even numbers give an even number?
2. Why does adding two odd numbers give an even number?
3. Why does adding an odd and an even number give an odd number?
4. Why aren’t there more even than odd numbers?
I’ll answer each one for you although the last is going to take the most explaining. *grin* So, here goes.
To start with, the answers to these questions are referring only to integers (…, -3, -2, -1, 0, 1, 2, 3, …).
There are several things to keep in mind.
When you add two integers, the answer is always another integer (this is a property of the set of integers called closure).
The associative property – for any integers m, n, p; m + n + p = (m + n) + p = m + (n + p).
The distributive property – for any integers a, b, c; ab + ac = a(b + c).
Now onto the first three questions.
1. Why does adding two even numbers give an even number?
Let’s start with a definition. We are defining an even number as a number that when divided by 2 has no remainder. We can write any even number as 2k where k is any integer. It doesn’t matter if it’s even or odd since multiplying by 2 immediately makes the result (2k) even.
Now add two even numbers. By the definition above, we can choose any two even numbers. Let’s call them 2k and 2m where k and m are integers.
Do the arithmetic: 2k + 2m = 2(k +m) by the distributive property. (k + m) is an integer because adding two integers always results in another integer. Since 2(k + m) is divisible by 2 with no remainder (see our original definition), adding 2k + 2m results in an even number.
2. Why does adding two odd numbers give an even number?
First we’ll define an odd number as a number that when divided by 2 results in a remainder of 1. We know that for any n = 2k, n is even from above. What we want is an n that is odd. By our definition of an odd number, n = 2k + 1 is odd because when we divide n by 2 we end up with a remainder of 1.
Now add two odd numbers. As above, we can choose any two odd numbers. Let’s choose n = 2k + 1 and m = 2r + 1 where k and r are integers.
Next the arithmetic: n + m = (2k + 1) + (2r + 1) = 2k + 2r + 2 (combining like terms).
2k + 2r + 2 = 2(k + r + 1) by the distributive property. Since k, r, and 1 are integers, (k + r + 1) is also an integer. So we now have an even number by our previous definition of an even number.
3. Why does adding an odd and an even number give an odd number?
Choose any odd number m = 2k + 1 (by our previous definition of an odd number) and any even number n = 2p (by our previous definition of an even number) where m, k, n, and p are integers. Now do the addition.
m + n = (2k + 1) + (2p) = 2k + 2p + 1 = (2k + 2p) + 1 (by the associative property) and
(2k + 2p) + 1 = 2(k + p) + 1 (by the distributive property). Since k and p are integers, k + p is an integer, and so we have by definition an odd number.
Now for the last question – why aren’t there more even than odd numbers?
I need to start by talking about infinite sets and how we tell their “size” or cardinality in math speak. The set of natural numbers (counting numbers – 1, 2, 3, …) is an infinite set. It’s a countably infinite set because (at least in theory) you could count all the members of the set. Of course, since there is no “largest” natural number, you’d never actually count to the end. Georg Cantor, when he described infinite sets, said that the set of natural numbers has a cardinality (size) of aleph-0. All countably infinite sets have this cardinality of aleph-0. So all countably infinite sets are the same size. Got that?
In order to show that a set is countably infinite, I have to find a way to match each member of the set to each member of the set of natural numbers. This is a one-to-one correspondence between the two sets. The good thing is that I can rearrange the numbers in the set I’m trying to put into one-to-one correspondence with the set of natural numbers – order, in the every day sense, doesn’t matter in this case.
Let’s start with the even numbers. The set of even numbers (I’m using integers) is (…, -6, -4, -2, 0, 2, 4, 6, …). This is a problem because the even numbers go on forever in the negative and in the positive direction. But I can reorder the set and end up with (0, -2, 2, -4, 4, -6, 6,...).
With this reordering, I can make the match to the set of natural numbers like this (my one-to-one correspondence) by lining up the natural numbers under the even numbers.
0 -2 2 -4 4 -6 6 and so forth (the even numbers) 1 2 3 4 5 6 7 and so forth (the natural numbers)
Since there is a one-to-one correspondence between the set of even numbers and the set of natural numbers, the set of even numbers is countably infinite and has the same cardinality (size) as the natural numbers – aleph-0.
The odd numbers work the same way.
Reorder the set of odd numbers (…, -5, -3, -1, 0, 1, 3, 5, …) to (0, -1, 1, -3, 3, -5, 5, …).
Now you can make a one-to-one correspondence to the natural numbers in the same way as for the even numbers. With this one-to-one correspondence, the set of odd numbers is countably infinite and so has a cardinality (size) of aleph-0.
Since the set of odd numbers and the set of even numbers both have a cardinality of aleph-0, they are the same size.
BTW – the set of rational numbers (your fractions – ½, ¼) is also countably infinite which makes it the same size as the set of integers, the set of even numbers, and the set of odd numbers. The set of real numbers (includes the integers, the rationals, and the irrational numbers) though, is uncountably infinite and has a different cardinality. Proofs are here. I think the proof for the rational numbers is pretty slick.
vw bug said...
You are wonderful. Thanks!!!
Harvey said...
Heh. That was actually pretty cool :-)
MathCogIdiocy said...
I knew I could find some math for all of you. *grin*
zapkid said...
Thats more then some stuff :P , i stopped reading the blogs when they were ban back here in pakistan , but know i think i should start it again
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# if | r | != 1 is integer r even 1 r is not positive 2 2r
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if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
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Updated on: 16 Apr 2014, 01:13
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Orginiallly Posted Question
if | r | != 1 is integer r even
1 r is not positive
2 2r > -5
OA is C but i am not convinced
CORRECT QUESTION
If r is an integer and $$|r|!=1$$ is $$r$$ even?
(1) $$r$$ is not positive.
(2) $$5r>-2$$.
Originally posted by rxs0005 on 14 Aug 2010, 04:38.
Last edited by WoundedTiger on 16 Apr 2014, 01:13, edited 1 time in total.
Question Formatting done
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Posts: 60647
### Show Tags
14 Aug 2010, 05:18
rxs0005 wrote:
if | r | != 1 is integer r even
1 r is not positive
2 2r > -5
OA is C but i am not convinced
First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.
$$|r|!=1$$ --> $$|r|=0$$ or $$|r|=1$$, so $$r$$ can take 3 values: 0, 1, and -1, out of which only 0 is even. So basicalyy questions asks: is $$r=0$$?
(1) $$r$$ is not positive --> $$r$$ can be 0 or -1. Not sufficient.
(2) $$2r>-5$$ --> $$r>-\frac{5}{2}=-2.5$$ --> $$r$$ can be 0, 1 or -1. Not sufficient.
(1)+(2) still two values of $$r$$ are possible 0 and -1. Not sufficient.
If statement (2) were $$5r>-2$$, then answer would be C: $$r>-\frac{2}{5}=-0.4$$ --> $$r$$ can be 0 or 1. Not sufficient.
(1)+(2) only one value of $$r$$ is possible: 0, hence $$r$$ is even. Sufficient.
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### Show Tags
14 Aug 2010, 07:53
Bunuel u mistook the statement
what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion
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Posts: 60647
### Show Tags
14 Aug 2010, 08:15
rxs0005 wrote:
Bunuel u mistook the statement
what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion
This means that I wrote new question:
If r is an integer and $$|r|!=1$$ is $$r$$ even?
(1) $$r$$ is not positive.
(2) $$5r>-2$$.
As for original question:
If $$|r|\neq{1}$$ is $$r=even$$?
$$|r|\neq{1}$$ --> $$r\neq{1}$$ and $$r\neq{-1}$$.
(1) $$r$$ is not positive --> Clearly insufficient, $$r$$ can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).
(2) $$2r>-5$$ --> $$r>-\frac{5}{2}=-2.5$$ --> again $$r$$ can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.
(1)+(2) $$r$$ is not positive and $$r>-2.5$$ --> $$r$$ can be -2, -1, or 0. But as given that $$r\neq{-1}$$ then only valid solutions for $$r$$ are -2 and 0, both are even. Sufficient.
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### Show Tags
14 Aug 2010, 08:33
Bunuel,
If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped?
and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E
I just saw, (2) was a typo. Very rare you make a mistake!
In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works.
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### Show Tags
14 Aug 2010, 08:55
true, the condition for statement 2 has been flipped.
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### Show Tags
14 Aug 2010, 09:07
mainhoon wrote:
Bunuel,
If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped?
and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E
I just saw, (2) was a typo. Very rare you make a mistake!
In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works.
No mistake there. Nothing was flipped. Look carefully, I consider in my previous post two questions:
MINE:
If r is an integer and $$|r|!=1$$ is $$r$$ even?
(1) $$r$$ is not positive.
(2) $$5r>-2$$.
AND ORIGINAL:
If $$|r|\neq{1}$$ is $$r=even$$?
(1) $$r$$ is not positive.
(2) $$2r>-5$$.
Totally in my previous posts we have 3 questions. First 2 I "invented" because rxs0005 meant by | r | != 1 $$|r|\neq{1}$$ and I understood it as it was written $$|r|!={1}$$. So two questions I consider in my first post are those I "invented".
Question 1 (mine):
If r is an integer and $$|r|!=1$$ is $$r$$ even?
First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.
$$|r|!=1$$ --> $$|r|=0$$ or $$|r|=1$$, so $$r$$ can take 3 values: 0, 1, and -1, out of which only 0 is even. So basically questions asks: is $$r=0$$?
(1) $$r$$ is not positive --> $$r$$ can be 0 or -1. Not sufficient.
(2) $$2r>-5$$ --> $$r>-\frac{5}{2}=-2.5$$ --> $$r$$ can be 0, 1 or -1. Not sufficient.
(1)+(2) still two values of $$r$$ are possible 0 and -1. Not sufficient.
Question #2 (mine):
If r is an integer and $$|r|!=1$$ is $$r$$ even?
(1) $$r$$ is not positive --> $$r$$ can be 0 or -1. Not sufficient.
(2) $$5r>-2$$ --> $$r>-\frac{2}{5}=-0.4$$ --> $$r$$ can be 0 or 1. Not sufficient.
(1)+(2) only one value of $$r$$ is possible: 0, hence $$r$$ is even. Sufficient.
Question #3 (original):
If $$|r|\neq{1}$$ is $$r=even$$?
$$|r|\neq{1}$$ --> $$r\neq{1}$$ and $$r\neq{-1}$$.
(1) $$r$$ is not positive --> Clearly insufficient, $$r$$ can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).
(2) $$2r>-5$$ --> $$r>-\frac{5}{2}=-2.5$$ --> again $$r$$ can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.
(1)+(2) $$r$$ is not positive and $$r>-2.5$$ --> $$r$$ can be -2, -1, or 0. But as given that $$r\neq{-1}$$ then only valid solutions for $$r$$ are -2 and 0, both are even. Sufficient.
BUT: if the second statement were as you wrote answer still would be C.
Question #4 (your's):
If $$|r|\neq{1}$$ is $$r=even$$?
$$|r|\neq{1}$$ --> $$r\neq{1}$$ and $$r\neq{-1}$$.
(1) $$r$$ is not positive --> Clearly insufficient, $$r$$ can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).
(2) $$5r>-2$$ --> $$r>-\frac{2}{5}=-0.4$$ --> $$r$$ can be 0 or 2, 3, 4, ... Not sufficient.
(1)+(2) Only one solution is possible for $$r$$ is 0, which is even. Sufficient.
Hope it's clear.
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### Show Tags
14 Aug 2010, 09:22
Bunuel wrote:
rxs0005 wrote:
Bunuel u mistook the statement
what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion
This means that I wrote new question:
If r is an integer and $$|r|!=1$$ is $$r$$ even?
(1) $$r$$ is not positive.
(2) $$5r>-2$$.
As for original question:
If $$|r|\neq{1}$$ is $$r=even$$?
$$|r|\neq{1}$$ --> $$r\neq{1}$$ and $$r\neq{-1}$$.
(1) $$r$$ is not positive --> Clearly insufficient, $$r$$ can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).
(2) $$2r>-5$$ --> $$r>-\frac{5}{2}=-2.5$$ --> again $$r$$ can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.
(1)+(2) $$r$$ is not positive and $$r>-2.5$$ --> $$r$$ can be -2, -1, or 0. But as given that $$r\neq{-1}$$ then only valid solutions for $$r$$ are -2 and 0, both are even. Sufficient.
Bunuel - I see that there was an original question and you had modified it. I got it. Thanks for the clarification. As I said, very rare you make a mistake
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
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15 Apr 2014, 14:01
OA on original question should be E.
Thanks
Cheers!
J
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
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16 Apr 2014, 01:14
jlgdr wrote:
OA on original question should be E.
Thanks
Cheers!
J
Question formatting done.
OA is correct
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
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22 Apr 2014, 06:57
jlgdr wrote:
OA on original question should be E.
Thanks
Cheers!
J
Hi jlgdr,
The answer for the original question is indeed E , if "!" meant as factorial. However if the same is mentioned as "not equal to", then the answer should be C. I believe the question is originally posted for "!" as nt equal to. In that case "C" is the correct answer as mentioned in the OA.
Hope that helps.
Thanks.
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink] 21 Dec 2017, 14:06
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Basic Math
Algebra, precalculus and calculus for college and university students. Contains topics ranging from numbers to differentiation and integration.
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Course content
Numbers
Integers
THEORY
T
1.
Integers
PRACTICE
P
2.
Integers
5
THEORY
T
3.
Ordering of integers
PRACTICE
P
4.
Ordering of integers
8
THEORY
T
5.
Sum, terms, product and factors
PRACTICE
P
6.
Sum, terms, product and factors
5
THEORY
T
7.
Divisors
PRACTICE
P
8.
Divisors
6
THEORY
T
9.
Order of operations
PRACTICE
P
10.
Order of operations
8
THEORY
T
11.
Factorization
PRACTICE
P
12.
Factorization
5
THEORY
T
13.
Prime numbers
PRACTICE
P
14.
Prime numbers
5
THEORY
T
15.
Prime factorization
PRACTICE
P
16.
Prime factorization
7
THEORY
T
17.
Greatest common divisor and least common multiple
PRACTICE
P
18.
Greatest common divisor and least common multiple
9
Negative numbers
THEORY
T
1.
PRACTICE
P
2.
8
THEORY
T
3.
Multiplying negative numbers
PRACTICE
P
4.
Multiplying negative numbers
8
THEORY
T
5.
Dividing negative numbers
PRACTICE
P
6.
Dividing negative numbers
6
THEORY
T
7.
Opposite numbers
PRACTICE
P
8.
Opposite numbers
5
THEORY
T
9.
Absolute value
PRACTICE
P
10.
Absolute value
6
Fractions
THEORY
T
1.
Fractions
PRACTICE
P
2.
Fractions
5
THEORY
T
3.
Equivalent fractions
PRACTICE
P
4.
Equivalent fractions
7
THEORY
T
5.
Negative fractions
PRACTICE
P
6.
Negative fractions
10
THEORY
T
7.
Simplifying Fractions
PRACTICE
P
8.
Simplifying Fractions
10
THEORY
T
9.
Adding and subtracting fractions with like denominators
PRACTICE
P
10.
Adding and subtracting fractions with like denominators
5
THEORY
T
11.
Writing fractions with like denominators
PRACTICE
P
12.
Writing fractions with like denominators
8
THEORY
T
13.
PRACTICE
P
14.
9
THEORY
T
15.
Convenient addition and subtraction of fractions
PRACTICE
P
16.
Convenient addition and subtraction of fractions
8
THEORY
T
17.
Multiplication of fractions
PRACTICE
P
18.
Multiplication of fractions
8
THEORY
T
19.
Reciprocal of a fraction
PRACTICE
P
20.
Reciprocal of a fraction
7
THEORY
T
21.
Division of fractions
PRACTICE
P
22.
Division of fractions
10
Powers and roots
THEORY
T
1.
Powers
PRACTICE
P
2.
Powers
6
THEORY
T
3.
Fractions raised to an integral power
PRACTICE
P
4.
Fractions raised to an integral power
5
THEORY
T
5.
Rules of calculation for powers
PRACTICE
P
6.
Rules of calculation for powers
8
THEORY
T
7.
Negative exponents
PRACTICE
P
8.
Negative exponents
7
THEORY
T
9.
Square roots
PRACTICE
P
10.
Square roots
6
THEORY
T
11.
Rules of calculation for roots
PRACTICE
P
12.
Rules of calculation for roots
5
THEORY
T
13.
Roots of fractions
PRACTICE
P
14.
Roots of fractions
5
THEORY
T
15.
Standard notation of roots
PRACTICE
P
16.
Standard form of roots
5
THEORY
T
17.
Higher roots
PRACTICE
P
18.
Higher roots
5
THEORY
T
19.
Rules of calculation for higher roots
PRACTICE
P
20.
Rules of calculation for higher roots
8
THEORY
T
21.
Standard notation of higher roots
PRACTICE
P
22.
Standard notation of higher roots
6
THEORY
T
23.
Order of operations with powers and roots
PRACTICE
P
24.
Order of operations with powers and roots
10
THEORY
T
25.
Irrational numbers
PRACTICE
P
26.
Irrational numbers
5
Ratios
THEORY
T
1.
Decimal numbers
PRACTICE
P
2.
Decimal numbers
5
THEORY
T
3.
Ordering of decimal numbers
PRACTICE
P
4.
Ordering of decimal numbers
8
THEORY
T
5.
Rounding numbers
PRACTICE
P
6.
Rounding numbers
10
THEORY
T
7.
Percentages
PRACTICE
P
8.
Percentages
8
THEORY
T
9.
Ratios
PRACTICE
P
10.
Ratios
7
THEORY
T
11.
Fractions, decimals, percentages, and ratios
PRACTICE
P
12.
Fractions, decimals, percentages, and ratios
7
Algebra
Variables
THEORY
T
1.
Variables
PRACTICE
P
2.
Variables
4
THEORY
T
3.
Sum and product of variables
PRACTICE
P
4.
Sum and product of variables
5
THEORY
T
5.
Substitution
PRACTICE
P
6.
Substitution
7
THEORY
T
7.
Simplification
PRACTICE
P
8.
Simplification
5
THEORY
T
9.
Simplification with algebraic rules
PRACTICE
P
10.
Simplification with algebraic rules
5
Calculating with exponents and roots
THEORY
T
1.
Integer exponents
PRACTICE
P
2.
Integer exponents
7
THEORY
T
3.
Calculating with integer exponents
PRACTICE
P
4.
Positive integer exponents 1
10
PRACTICE
P
5.
Positive integer exponents 2
10
PRACTICE
P
6.
Positive integer exponents 3
7
PRACTICE
P
7.
Positive integer exponents 4
6
THEORY
T
8.
Square roots
PRACTICE
P
9.
Square roots
5
THEORY
T
10.
Calculating with square roots
PRACTICE
P
11.
Calculating with square roots
5
THEORY
T
12.
Higher degree roots
PRACTICE
P
13.
Higher degree roots
8
THEORY
T
14.
Calculating with fractional exponents
PRACTICE
P
15.
Calculating with fractional exponents
7
THEORY
T
16.
Order of operations
PRACTICE
P
17.
Order of operations
5
Expanding brackets
THEORY
T
1.
Expanding brackets
PRACTICE
P
2.
Expanding brackets 1
9
PRACTICE
P
3.
Expanding brackets 2
5
THEORY
T
4.
Expanding double brackets
PRACTICE
P
5.
Expanding double brackets 1
10
PRACTICE
P
6.
Expanding double brackets 2
10
PRACTICE
P
7.
Expanding double brackets 3
6
Factorization
THEORY
T
1.
Factoring out
PRACTICE
P
2.
Factoring out
8
THEORY
T
3.
Factorization
PRACTICE
P
4.
Factorization 2
7
PRACTICE
P
5.
Factorization 3
10
PRACTICE
P
6.
Factorization 4
6
Notable Products
THEORY
T
1.
The square of a sum or a difference
PRACTICE
P
2.
The square of a sum or a difference
7
THEORY
T
3.
The difference of two squares
PRACTICE
P
4.
The difference of two squares
8
THEORY
T
1.
Fractions
PRACTICE
P
2.
Fractions
5
THEORY
T
3.
Simplifying fractions
PRACTICE
P
4.
Simplifying fractions
3
PRACTICE
P
5.
Simplifying fractions
5
THEORY
T
6.
Addition and subtraction of like fractions
PRACTICE
P
7.
Addition and subtraction of like fractions
5
THEORY
T
8.
Making fractions similar
PRACTICE
P
9.
Making fractions similar
5
THEORY
T
10.
PRACTICE
P
11.
8
THEORY
T
12.
Multiplication of fractions
PRACTICE
P
13.
Multiplication of fractions
8
THEORY
T
14.
Division of fractions
PRACTICE
P
15.
Division of fractions
6
THEORY
T
16.
Fraction decomposition
PRACTICE
P
17.
Fraction decomposition
5
Linear formulas and equations
Formulas
THEORY
T
1.
Formula
PRACTICE
P
2.
Formulas
6
THEORY
T
3.
Dependent and independent variables
PRACTICE
P
4.
Dependent and independent variables
5
THEORY
T
5.
Graphs
PRACTICE
P
6.
Graphs
8
Linear functions
THEORY
T
1.
Linear formula
PRACTICE
P
2.
Linear formula
6
THEORY
T
3.
Slope and intercept
PRACTICE
P
4.
Slope and intercept
6
THEORY
T
5.
Composing a linear formula
PRACTICE
P
6.
Composing a linear formula
5
THEORY
T
7.
Parallel and intersecting linear formulas
PRACTICE
P
8.
Parallel and intersecting linear formulas
6
Linear equations and inequalities
THEORY
T
1.
Linear equations
PRACTICE
P
2.
Linear equations
12
THEORY
T
3.
The general solution of a linear equation
PRACTICE
P
4.
The general solution of a linear equation
10
THEORY
T
5.
Intersection points of linear formulas with the axes
PRACTICE
P
6.
Intersection points of linear formulas with the axes
7
THEORY
T
7.
Intersection point of two linear formulas
PRACTICE
P
8.
Intersection point of two linear formulas
10
THEORY
T
9.
Linear inequalities
PRACTICE
P
10.
Linear inequalities
7
THEORY
T
11.
General solution of a linear inequality
PRACTICE
P
12.
General solution of a linear inequality
10
Systems of linear equations
An equation of a line
THEORY
T
1.
A linear equation with two unknowns
PRACTICE
P
2.
A linear equation with two unknowns
5
THEORY
T
3.
Solution linear equation with two unknowns
PRACTICE
P
4.
Solution linear equation with two unknowns
5
THEORY
T
5.
The equation of a line
PRACTICE
P
6.
The equation of a line
10
THEORY
T
7.
Composing the equation of a line
PRACTICE
P
8.
Composing the equation of a line
7
Two equations with two unknowns
THEORY
T
1.
Systems of linear equations
PRACTICE
P
2.
Systems of linear equations
5
THEORY
T
3.
Solving systems of linear equations by substitution
PRACTICE
P
4.
Solving systems of linear equations by substitution
10
THEORY
T
5.
Solving systems of equations by elimination
PRACTICE
P
6.
Solving systems of equations by elimination
5
THEORY
T
7.
General solution system of linear equations
PRACTICE
P
8.
General solution system of linear equations
9
Parabola
THEORY
T
1.
PRACTICE
P
2.
8
THEORY
T
3.
Parabola
PRACTICE
P
4.
Parabola
6
THEORY
T
1.
PRACTICE
P
2.
5
THEORY
T
3.
PRACTICE
P
4.
10
THEORY
T
5.
Solving quadratic equations by completing the square
PRACTICE
P
6.
Solving quadratic equations by completing the square
5
THEORY
T
7.
PRACTICE
P
8.
6
PRACTICE
P
9.
10
Drawing parabolas
THEORY
T
1.
Intersection of parabolas with the axes
PRACTICE
P
2.
Intersections of parabolas with the axes
10
THEORY
T
3.
Vertex of a parabola
PRACTICE
P
4.
Vertex of a parabola
5
THEORY
T
5.
Drawing of parabolas
PRACTICE
P
6.
Drawing of parabolas
16
THEORY
T
7.
Transformations of parabolas
PRACTICE
P
8.
Transformations of parabolas
8
Intersection points of parabolas
THEORY
T
1.
Intersection points of a parabola with a line
PRACTICE
P
2.
Intersection points of parabolas 1
8
THEORY
T
3.
Intersection points of parabolas
PRACTICE
P
4.
Intersection points of parabolas
11
THEORY
T
1.
PRACTICE
P
2.
8
Functions
Domain and range
THEORY
T
1.
Function and formula
PRACTICE
P
2.
Function and formula
6
THEORY
T
3.
Function rule
PRACTICE
P
4.
Function rule
5
THEORY
T
5.
Intervals
PRACTICE
P
6.
Intervals
7
THEORY
T
7.
Domain
PRACTICE
P
8.
Domain
2
THEORY
T
9.
Range
PRACTICE
P
10.
Range
2
Power functions
THEORY
T
1.
Power functions
PRACTICE
P
2.
Power functions
5
THEORY
T
3.
Transformations of power functions
PRACTICE
P
4.
Transformations of power functions
12
THEORY
T
5.
Equations with power functions
PRACTICE
P
6.
Equations with power functions
5
Higher degree polynomials
THEORY
T
1.
Polynomials
PRACTICE
P
2.
Polynomials
3
THEORY
T
3.
Equations with polynomials
PRACTICE
P
4.
Equations with polynomials
4
THEORY
T
5.
Solving higher degree polynomials with factorization
PRACTICE
P
6.
Solving higher degree polynomials with factorization
4
THEORY
T
7.
Solving higher degree polynomials with the quadratic equation
PRACTICE
P
8.
Solving higher degree polynomials with the quadratic equation
2
THEORY
T
9.
Higher degree inequalities
PRACTICE
P
10.
Higher degree inequalities
8
Power functions and root functions
THEORY
T
1.
Root function
PRACTICE
P
2.
Root functions
2
THEORY
T
3.
Transformations of root functions
PRACTICE
P
4.
Transformations of root functions
5
THEORY
T
5.
Root equations
PRACTICE
P
6.
Root equations
2
THEORY
T
7.
Solving root equations with substitution
PRACTICE
P
8.
Solving root equations with substitution
2
THEORY
T
9.
Inverse functions
PRACTICE
P
10.
Inverse functions
2
Fractional functions
THEORY
T
1.
Asymptotes and hyperbolas
PRACTICE
P
2.
Asymptotes and hyperbolas
5
THEORY
T
3.
Power functions with negative exponents
PRACTICE
P
4.
Power functions with negative exponents
2
THEORY
T
5.
Transformations of power functions with negative exponents
PRACTICE
P
6.
Transformations of power functions with negative exponents
12
THEORY
T
7.
Linear fractional functions
PRACTICE
P
8.
Linear fractional functions
7
THEORY
T
9.
Linear fractional equations
PRACTICE
P
10.
Linear fractional equations
3
THEORY
T
11.
Inverse of linear fractional function
PRACTICE
P
12.
Inverse of linear fractional function
4
THEORY
T
13.
Quotient functions
PRACTICE
P
14.
Quotient functions
16
Exponential functions and logarithms
Exponential functions
THEORY
T
1.
The exponential function
PRACTICE
P
2.
The exponential function
4
THEORY
T
3.
Exponential equations
PRACTICE
P
4.
Exponential equations
5
THEORY
T
5.
Transformations of the exponential function
PRACTICE
P
6.
Transformations of the exponential function
2
Logarithmic functions
THEORY
T
1.
The logarithmic function
PRACTICE
P
2.
The logarithmic function
9
THEORY
T
3.
Logarithmic equations
PRACTICE
P
4.
Logarithmic equations
5
THEORY
T
5.
Exponential equations
PRACTICE
P
6.
Exponential equations
5
THEORY
T
7.
Isolating variables
PRACTICE
P
8.
Isolating variables
6
THEORY
T
9.
Rules for logarithms
PRACTICE
P
10.
Rules for logarithms
5
THEORY
T
11.
More logarithmic equations
PRACTICE
P
12.
More logarithmic equations
5
THEORY
T
13.
Change of base
PRACTICE
P
14.
Change of base
5
THEORY
T
15.
Solving equations using substitution
PRACTICE
P
16.
Solving equations using substitution
5
THEORY
T
17.
Graph of logarithmic function
PRACTICE
P
18.
Graph of logarithmic function
2
THEORY
T
19.
Transformations of the logarithmic function
PRACTICE
P
20.
Transformations of the logarithmic function
2
Trigonometry
Angles with sine, cosine and tangent
THEORY
T
1.
Angles
PRACTICE
P
2.
Angles
1
THEORY
T
3.
Triangles
PRACTICE
P
4.
Triangles
1
THEORY
T
5.
Rules for right-angled triangles
PRACTICE
P
6.
Rules for right-angled triangles 1
9
PRACTICE
P
7.
Rules for right-angled triangles 2
6
THEORY
T
8.
PRACTICE
P
9.
8
THEORY
T
10.
Symmetry in the unit circle
PRACTICE
P
11.
Symmetry in the unit circle
10
THEORY
T
12.
Special values of trigonometric functions
PRACTICE
P
13.
Special values of trigonometric functions
5
THEORY
T
14.
PRACTICE
P
15.
7
THEORY
T
16.
Sine and cosine rules
PRACTICE
P
17.
Sine and cosine rules
8
Trigonometric functions
THEORY
T
1.
Trigonometric functions
PRACTICE
P
2.
Trigonometric functions
4
THEORY
T
3.
Transformations of trigonometric functions
PRACTICE
P
4.
Transformations of trigonometric functions
15
THEORY
T
5.
Inverse trigonometric functions
PRACTICE
P
6.
Inverse trigonometric functions
4
THEORY
T
7.
Trigonometric equations 1
PRACTICE
P
8.
Trigonometric equations 1
6
THEORY
T
9.
Trigonometric equations 2
PRACTICE
P
10.
Trigonometric equations 2
6
Differentiation
The derivative
THEORY
T
1.
The difference quotient
PRACTICE
P
2.
The difference quotient
5
THEORY
T
3.
The difference quotient at a point
PRACTICE
P
4.
The difference quotient at a point
5
THEORY
T
5.
The tangent line
PRACTICE
P
6.
The tangent line
2
THEORY
T
7.
The notion of derivative
PRACTICE
P
8.
The notion of derivative
14
The derivative of power functions
THEORY
T
1.
The derivative of power functions
PRACTICE
P
2.
The derivative of power functions 1
7
PRACTICE
P
3.
The derivative of power functions 2
8
PRACTICE
P
4.
The derivative of power functions 3
5
Sum and product rule
THEORY
T
1.
The sum rule
PRACTICE
P
2.
The sum rule
7
THEORY
T
3.
The product rule
PRACTICE
P
4.
The product rule
6
Chain rule
THEORY
T
1.
Composite functions
PRACTICE
P
2.
Composite functions
8
THEORY
T
3.
The chain rule
PRACTICE
P
4.
The chain rule
9
The derivative of standard functions
THEORY
T
1.
The derivative of trigonometric functions
PRACTICE
P
2.
The derivative of trigonometric functions 1
8
PRACTICE
P
3.
The derivative of trigonometric functions 2
10
THEORY
T
4.
The base e and the natural logarithm
PRACTICE
P
5.
The base e and the natural logarithm
5
THEORY
T
6.
The derivative of exponential functions and logarithms
PRACTICE
P
7.
The derivative of exponential functions and logarithms 1
7
Quotient rule
THEORY
T
1.
The quotient rule
PRACTICE
P
2.
The quotient rule 1
6
PRACTICE
P
3.
The quotient rule 2
7
Mixed exercises differentiation
PRACTICE
P
1.
Mixed exercises differentiation
15
Applications of derivatives
THEORY
T
1.
Increasing and decreasing
PRACTICE
P
2.
Increasing and decreasing
8
THEORY
T
3.
Extreme values
PRACTICE
P
4.
Extreme values
9
THEORY
T
5.
The second derivative
PRACTICE
P
6.
The second derivative
5
THEORY
T
7.
Types of increasing and decreasing
PRACTICE
P
8.
Types of increasing and decreasing
5
THEORY
T
9.
Inflection points
PRACTICE
P
10.
Inflection points
5
THEORY
T
11.
Higher order derivatives
PRACTICE
P
12.
Higher order derivatives
5
Integration
Antiderivatives
THEORY
T
1.
The antiderivative of a function
PRACTICE
P
2.
The antiderivative of a function
5
THEORY
T
3.
The antiderivative of a power function
PRACTICE
P
4.
The antiderivative of a power function
5
THEORY
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# C1 Differentiation question [SOLVED]Watch
Announcements
#1
A curve has equation y = x3/2 + 48x-1/2
a) Find the derivative of y with respect to x
b) Find the equation of the tangent to this curve at the point where x=1
I think I could do b easily but I don't understand part a.
1
4 years ago
#2
A curve has equation y = x3/2 + 48x-1/2
a) Find the derivative of y with respect to x
b) Find the equation of the tangent to this curve at the point where x=1
I think I could do b easily but I don't understand part a.
Can you do Part B, then? You might find Part B's answer useful to doing Part A.
1
4 years ago
#3
Part a) dy/dx = (times by power, then lower the power).
Part b) If I'm not mistaken (i haven't done this in a while) you simply use: y-y1 = m(x-x1)
Where x1 is the x value given, y1 is the y value when x = 1 (sub x=1 into the original equation). m is the gradient, sub x=1 into your part a Answer.
I hope this helps.
0
4 years ago
#4
A curve has equation y = x3/2 + 48x-1/2
a) Find the derivative of y with respect to x
b) Find the equation of the tangent to this curve at the point where x=1
I think I could do b easily but I don't understand part a.
Hi,
I hope this helps, it should be correct. Check wolfram alpha or something to be sure.
0
4 years ago
#5
A curve has equation y = x3/2 + 48x-1/2
a) Find the derivative of y with respect to x
b) Find the equation of the tangent to this curve at the point where x=1
I think I could do b easily but I don't understand part a.
Yeah, the workings are correct - hope it helps!
0
4 years ago
#6
For Part A, it's asking you to find the first derivative of the equation (dy/dx). To do this, just multiply the coefficient of each term by the power that x is raised to, and then take away 1 from that power (for example, if y=2x2-3x2/3, then dy/dx=4x-2x-1/3).
Part B can't be done without Part A, as you need to find the equation of the line via (y-y1)=m(x-x1), where m is the gradient and (x1,y1) is a point on the line. As you already know x1 (it is given that x=1), you can find y1 by substituting x=1 into the original equation. To find the gradient, m, substitute x=1 into your answer for Part A, then sub all those values into
(y-y1)=m(x-x1) and simplify to find the equation of the line.
0
#7
Thank you so much everyone!
0
4 years ago
#8
(Original post by Willis_l96)
*
You should not post full solutions.
0
4 years ago
#9
(Original post by keromedic)
You should not post full solutions.
I would usually agree, but this is Core 1 lmao
0
4 years ago
#10
(Original post by Willis_l96)
I would usually agree, but this is Core 1 lmao
Which means the OP is even less experience and would probably benefit more from being guided to the answer.
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# RATT RACE: Investing in resources such as employees
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04 May 2016, 08:51
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Investing in resources such as employees and technology is often more profitable over the long term than is to try to cut costs in these areas.
(A) is often more profitable over the long term than is to try to cut
(B) are often part of a more profitable long-term strategy than would be trying to cut
(C) is often a more profitable strategy over the long term than is trying to cut
(D) often leads to greater profits over the long term than does trying and cutting
(E) result often in larger profits than cutbacks in these areas
Day 14 Question of the Verbal Contest:GMAT Club RATT Race
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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04 May 2016, 09:06
Seems C to me.
Investing in resources such as employees and technology is often more profitable over the long term than is to try to cut costs in these areas.
(A) is often more profitable over the long term than is to try to cut
---Investing and to try is not parallel
(B) are often part of a more profitable long-term strategy than would be trying to cut
----Investing is singular
(C) is often a more profitable strategy over the long term than is trying to cut
(D) often leads to greater profits over the long term than does trying and cutting
--- trying and cutting is wrong.
(E) result often in larger profits than cutbacks in these areas
---investing and cutbacks are not parallel.
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RATT RACE: Investing in resources such as employees [#permalink]
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04 May 2016, 10:05
souvik101990 wrote:
Investing in resources such as employees and technology is often more profitable over the long term than is to try to cut costs in these areas.
(A) is often more profitable over the long term than is to try to cut
(B) are often part of a more profitable long-term strategy than would be trying to cut
(C) is often a more profitable strategy over the long term than is trying to cut
(D) often leads to greater profits over the long term than does trying and cutting
(E) result often in larger profits than cutbacks in these areas
Day 14 Question of the Verbal Contest:GMAT Club RATT Race
With Vikas10
IMHO (C) as well
Errors according to me are highlighted in red..
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RATT RACE: Investing in resources such as employees [#permalink]
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04 May 2016, 21:02
Its a comparison question and between option C and D, which seem correct in comparing "subject - subject" aspect, at first I thought C changes the meaning slightly by mentioning the "profitable strategy" instead of profits but thats not a deciding factor. Option D seems correct but there is an error in verb number "does" when it mentioned two activities - trying and cutting. Hence option C is the best answer.
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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05 May 2016, 07:02
Definitely C
First step should be to eliminate all answers except A and C as "Investing" is a singular act that should be represented by "is". Then eliminate A because "Investing....try" is not parallel. Answer C correctly uses singular form "is" and "trying" is parallel to "Investing".
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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05 May 2016, 10:41
C it is .. well explained in posts above. Basically, it boils down to SVA and parallelism.
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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05 May 2016, 12:53
2
In the given sentence, a comparison is made using the construction “more . . . than,” and comparisons need to be between things of the same type expressed in the same grammatical form. This sentence is comparing “[i]nvesting” with “to try.” “Investing” is a gerund (the -ing form of a verb used as a noun), while “to try” is an infinitive (the to form of a verb, in this case also being used as a noun). These terms are not in the same form, so this error needs to be fixed. Eliminate choice (A).
One way to analyze the remaining answer choices is to look at the verbs at the beginning. Though subtler than in some questions, a 3-2 split exists between a singular and plural verb. The subject of the verb is “Investing.” (Recall that the subject of a sentence never resides in a prepositional phrase, which is what “in resources such as employees and technology” is.) “Investing” is singular, so answer choice (B), with “are,” and choice (E), with “result,” are incorrect. Choice (E) also contains an illogical comparison and alters the meaning of the sentence, as its construction indicates that “profits” are larger than “cutbacks.” Eliminate (B) and (E).
Choices (C) and (D) both use a singular verb at the outset and have no errors until after the word “than.” Here, choice (D) uses the wrong idiom. The correct phrase is “try[ing] to [do],” not “try[ing] and [doing].” Answer choice (C) is correct.
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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05 May 2016, 12:59
Good comparison question. OE is explained thoroughly by souvik101990
Gerund vs infinitive
Investing in resources such as employees and technology is often more profitable over the long term than is to try to cut costs in these areas.
(A) is often more profitable over the long term than is to try to cut
(B) are often part of a more profitable long-term strategy than would be trying to cut
(C) is often a more profitable strategy over the long term than is trying to cut
(D) often leads to greater profits over the long term than does trying and cutting
(E) result often in larger profits than cutbacks in these areas
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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09 May 2016, 08:52
Well done by all who commented on this question! Our team at Kaplan wishes you the best in your continued studies!
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Re: RATT RACE: Investing in resources such as employees [#permalink]
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04 May 2017, 15:58
without doubt
investing is singular - therefore we need a verb in singular form.
investing is more profitable than cutting costs is.
only C fits
Re: RATT RACE: Investing in resources such as employees &nbs [#permalink] 04 May 2017, 15:58
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Question
# ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.
Open in App
Solution
## Finding the angles of cyclic quadrilateral:Step $1:$Finding the values of $\mathrm{x}$ and $\mathrm{y}$,In cyclic quadrilateral, the sum of the opposite angles is ${180}^{0}$.From figure, we get, $\angle \mathrm{A}=4\mathrm{y}+20\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=3\mathrm{y}-5\phantom{\rule{0ex}{0ex}}\angle \mathrm{C}=-4\mathrm{x}\phantom{\rule{0ex}{0ex}}\angle \mathrm{D}=-7\mathrm{x}+5$Therefore,$\angle C+\angle A={180}^{\circ }$$\begin{array}{rcl}4\mathrm{y}+20-4\mathrm{x}& =& 180\\ -4\mathrm{x}+4\mathrm{y}& =& 160\end{array}$ $\begin{array}{rcl}\mathrm{y}-\mathrm{x}& =& 40\end{array}$ …………………….$\left(1\right)$And, $\angle \mathrm{B}+\angle \mathrm{D}={180}^{\circ }$$\begin{array}{rcl}3\mathrm{y}-5-7\mathrm{x}+5& =& 180\end{array}$ $\begin{array}{rcl}3\mathrm{y}-7\mathrm{x}& =& 180\end{array}$ ……………….$\left(2\right)$On multiplying $3$ to equation $\left(1\right)$, we get,$\begin{array}{rcl}3\mathrm{y}-3\mathrm{x}& =& 120\end{array}$……………………………$\left(3\right)$On subtracting equation $\left(2\right)$ from equation $\left(3\right)$, we get$\begin{array}{rcl}3\mathrm{y}-3\mathrm{x}-\left(3\mathrm{y}-7\mathrm{x}\right)& =& 120-180\\ 3\mathrm{y}-3\mathrm{x}-3\mathrm{y}+7\mathrm{x}& =& -60\\ 4\mathrm{x}& =& -60\\ \mathrm{x}& =& -15\end{array}$Substituting this value in equation $\left(1\right)$, we get,$\begin{array}{rcl}\mathrm{y}-\mathrm{x}& =& 40\\ y-\left(-15\right)& =& 40\\ y+15& =& 40\\ y& =& 25\end{array}$Step $2:$ Finding the angles :On substituting the values of $\mathrm{x}$ and $\mathrm{y}$, we get,$\angle \mathrm{A}=4\mathrm{y}+20=4×25+20={120}^{\circ }\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=3\mathrm{y}-5=3×25-5={70}^{\circ }\phantom{\rule{0ex}{0ex}}\angle \mathrm{C}=-4\mathrm{x}=-4×\left(-15\right)={60}^{\circ }\phantom{\rule{0ex}{0ex}}\angle \mathrm{D}=-7\mathrm{x}+5=-7×\left(-15\right)+5={110}^{\circ }$Hence, the angles of the cyclic quadrilateral are $\angle A={120}^{\circ },\angle B={70}^{\circ },\angle C={60}^{\circ },\angle D={110}^{\circ }$.
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# Standard Deviation SD: Understanding Its Impact on Financial Analysis
✅ All InspiredEconomist articles and guides have been fact-checked and reviewed for accuracy. Please refer to our editorial policy for additional information.
## Standard Deviation Definition
Standard Deviation (SD) is a statistical measure that quantifies the amount of variation or dispersion in a set of values. A low SD indicates that the values tend to be close to the mean (average) value, while a high SD indicates that the values are spread out over a wider range.
## Understanding the Calculation of Standard Deviation SD
The actual process of calculating the standard deviation SD involves a few key steps, whether you’re dealing with data from a population or a sample.
### Calculating Population SD
First, let’s focus on how to compute standard deviation in the context of a population dataset. The steps involve:
1. Determine the mean (average) of the population.
2. Subtract the mean from each data point and square the result. These are your squared deviations.
3. Add up all the squared deviations.
4. Divide this by the total number of data points in your population. This is your variance.
5. Finally, take the square root of the variance to get the standard deviation.
The formula to calculate the population SD is:
``````SD = sqrt(Σ(x_i - μ)² / N)
``````
Where:
• SD: Standard Deviation
• x_i: Each individual data point
• μ: Population mean
• N: Number of data points in the population
### Calculating Sample SD
On the other hand, if you’re working with a smaller sample from a larger population, the calculation varies slightly:
1. Determine the sample mean.
2. Subtract the sample mean from each data point and square the result.
3. Add up all these squared deviations.
4. Divide by the number of sample data points minus 1. This is your sample variance.
5. Take the square root of the sample variance to get your sample standard deviation.
The reason we subtract 1, known as applying `(n-1)` or Bessel’s correction, is to correct for any bias in the estimation of the population variance and SD from a sample.
The formula to calculate the sample SD is:
``````SD_sample = sqrt(Σ(x_i - X_bar)² / (n-1))
``````
Where:
• SD_sample: Sample Standard Deviation
• x_i: Each individual data point
• X_bar: Sample mean
• n: Number of data points in the sample
The mean is of paramount importance in both these calculations as it represents the central or ‘average’ value of your data. The SD, then, gives insight into ‘how much’ individual data points deviate from this average.
Variations from the mean reveal the ‘volatility’ or ‘risk’ in your data set. The higher the standard deviation, the greater your data points stray from the mean. Conversely, a low SD indicates data points cluster closely around the mean. This information is incredibly valuable across numerous fields, from finance and economics, to demographics and public health.
## Applications of Standard Deviation in Finance
In finance, standard deviation serves a critical role in a wide range of applications. Here, we take a look at some of the ways finance professionals use it to gauge risk, optimize portfolios, price options, and construct volatility surfaces.
### Risk Measurement Tool
As a measure of variability or volatility, standard deviation is frequently used by financial analysts to assess investment risk. For instance, an equity with a high standard deviation has had a broad range of prices historically, indicating greater price volatility, and consequently, potentially higher risk. Analysts might use this measure to make decisions about investment allocations and risk management.
### Portfolio Optimization
In the realm of portfolio construction and management, standard deviation serves as a vital diagnostic tool. Portfolio managers use standard deviation to examine the expected risk or volatility of a portfolio of investments. By understanding variability, they can better estimate the portfolio’s overall risk and can optimize allocation to achieve better risk-adjusted returns. More specifically, the standard deviation of portfolio returns helps in measuring its riskiness and, in turn, supports the design of the optimal investment mix between high-risk, high-reward and low-risk, low-payout assets.
### Options Pricing
Standard deviation also plays a role in options pricing. In the Black-Scholes model – a widely used model for pricing options – standard deviation of the underlying asset returns is a key input. This value, referred to as volatility, directly influences the price of the option. A higher standard deviation, meaning more unpredictability in the asset’s price movement, will yield a higher price for the option, all else equal.
### Volatility Surface Construction
The concept of volatility surface comes into picture in the derivative market, specifically for pricing and trading of options. The standard deviation of asset returns is used to create a volatility surface, which plots the implied volatility of an option for various strike prices and expiry dates. This surface is then used by options traders to assess whether an option is underpriced or overpriced based on its implied volatility.
In sum, the application of standard deviation in finance is quite profound, aiding in measuring investment risk, designing optimised portfolios, pricing options, and constructing volatility surfaces. By quantifying volatility, standard deviation equips financial analysts with essential insights for decision-making and optimisation.
## The Role of Standard Deviation in Modern Portfolio Theory
As a critical element of Modern Portfolio Theory (MPT), standard deviation serves as a statistical measure of portfolio risk. The central concept of MPT is that of diversification, suggesting that investors can minimize risk in their portfolios by constructing an optimal mix of different types of assets that generate various levels of returns.
### The Importance of Diversification
Diversification is essentially the process of spreading your investment across a variety of assets to avoid concentrating risk in one area (e.g., one specific stock or asset class). The proverb “Don’t put all your eggs in one basket” encapsulates the essence of this principle.
What makes diversification useful is the inherent unpredictability of financial markets. By investing in a range of different assets, investors can protect themselves against substantial losses should the value of an individual asset plummet. Each asset’s performance in a portfolio will contribute to the portfolio’s overall risk and reward quotient. Therefore, standard deviation plays an invaluable role in understanding how the performance of each asset affects the portfolio’s total risk.
### The Efficient Frontier
Within the MPT framework, standard deviation helps investors visualize how diversification affects portfolio risk and return. This visualization explains the concept of the efficient frontier—an imaginary line in a risk-return graph representing optimum portfolios.
The efficient frontier is the set of optimal portfolios that offer the highest expected return for a defined level of risk. In other words, beyond this frontier, investors would experience increased risk without the likelihood of higher returns, thus defying the logic of diversification. Portfolios that align on the efficient frontier exhibit the lowest possible standard deviation without compromising on returns, hence they represent the most efficient deployment of an investor’s capital.
Therefore, within the context of Modern Portfolio Theory, the standard deviation isn’t just a mathematical concept. Rather, it’s a powerful tool that aids investors in understanding and quantifying risk, thus paving the way for portfolio optimization, efficient capital allocation, and ultimately, more robust financial decision-making.
## Interpretation and Misinterpretation of Standard Deviation
One approach to interpreting the standard deviation is to use the empirical rule, which states that for a normal distribution, almost all data will fall within three standard deviations of the mean. Practically speaking, this means if you find standard deviation value is relatively small, there’s less variation, and most data points are close to the mean. Conversely, a larger standard deviation value indicates more data variability, and data points are more spread away from the mean.
Misinterpretations often arise when people equate a high standard deviation with ‘bad’ and a low standard deviation with ‘good’. It’s problematic because these values by themselves provide no value judgement; they’re merely measures of dispersion. For example, a high standard deviation might indicate that there’s a high diversity in the data, which can be a positive aspect depending on the context.
The use of standard deviation isn’t without limitation. For instance, standard deviation is susceptible to outliers. A single outlier can significantly increase the calculated value of standard deviation, making the variability appear larger than it genuinely is.
Another limitation is that standard deviation only measures the variability around the mean. This implies that it is most appropriately used with symmetric distributions, or data sets that are roughly bell-shaped. But in case of skewed distributions, standard deviation can give misleading representation of variability.
Highlighting the use of standard deviation in isolation, it’s not recommended. Since standard deviation is based on the mean, it carries the same risks inherent to using the mean as a measure of central tendency. For example, the mean doesn’t always accurately reflect the ‘central’ or ‘typical’ value in a dataset, particularly in the presence of outliers. Thus, to get a reliable interpretation of data, it’s recommended to use standard deviation in conjunction with other statistical measures such as the mean, median, and mode.
Bear in mind also that standard deviation doesn’t directly indicate which observations could be considered ‘abnormal’ or ‘outliers’. Merely being more than one standard deviation from the mean doesn’t necessarily imply unusualness.
## Standard Deviation in Financial Modeling
In financial modeling and forecasting, standard deviation provides a measure of volatility and risk. A financial model often uses historical data on a given security or portfolio to compute the expected return and the expected standard deviation. These calculated values then paint a picture of the potential risk and reward.
The standard deviation comes in handy especially when modeling the future behavior of financially driven businesses, such as insurance companies and banks. While the expected return gives us an average outcome, a low standard deviation suggests that real-world outcomes will cluster closely around the average. Conversely, a high standard deviation implies greater potential volatility, with outcomes likely to vary more widely.
### Use in Monte Carlo Simulations
Monte Carlo simulations rely heavily on the standard deviation for its probabilistic input. These simulations involve running multiple trials with random inputs to project possible outcomes. The standard deviation helps define the range of possible investment returns, from the most optimistic to the most pessimistic scenarios.
Within a Monte Carlo method, the larger the standard deviation, the greater the likelihood of a wider array of potential outcomes. Such distributions provide a comprehensive view of the risk characteristics of investments, allowing analysts and investors to make more informed decisions about portfolio construction and risk management.
### Use in Risk Modeling
In risk modeling, the standard deviation again plays a crucial role. It helps quantify the uncertainty or the risk associated with a particular investment or a portfolio. If the standard deviation is high, it means the returns could be spread out over a large range of values. This implies a higher level of risk. On the other hand, a smaller standard deviation indicates that the returns are likely to be closer to the mean, suggesting lower risk.
Risk managers often use the standard deviation in combination with other mathematical tools to build models that can assess and predict various types of risks, such as market risk, credit risk, and operational risk. These models contribute to managing the overall investment risk and optimizing the risk-return trade-off.
### Use in Predictive Analytics
In predictive analytics, the standard deviation is used to estimate errors in models or forecasts. It provides a straightforward way of understanding the dispersion or the degree of spread of dataset values.
An analytical model with a low standard deviation suggests that the predicted values are close to the actual ones, indicating the model is reliable. Conversely, a high standard deviation suggests substantial potential divergence between projected and actual figures, potentially undermining confidence in the predictions.
## Standard Deviation and Corporate Social Responsibility (CSR)
The application of standard deviation in analyzing Corporate Social Responsibility (CSR) performance hinges on the assumption that financial indicators can reflect a company’s sustainability endeavors. Various financial indicators might be used in this analysis, such as profits, revenues, and operating costs. Standard deviation, as a measure of range and volatility, provides insight into how consistently these indicators function within a corporation.
### Measuring Sustainability with Standard Deviation
When assessing the standard deviation of these financial indicators, higher values indicate a higher degree of volatility. This volatility might be due to a variety of factors; however, in relation to CSR performance, it may indicate instability or inconsistency in a company’s sustainability practices. For example, a high standard deviation in yearly profits can suggest instability in a company’s income, which might be due to fluctuating investment in CSR initiatives.
A company focusing on sustainable practices will ideally have a consistent expenditure on CSR efforts. This should result in a lower standard deviation in related financial indicators, demonstrating consistency in financing and prioritizing CSR initiatives.
### Standard Deviation as an Indicator of Risk
Moreover, from an investment standpoint, standard deviation also serves as a risk profile indicator. A high standard deviation might indicate high volatility and potential financial risk, a trait that might dissuade investors conscious about CSR issues. Businesses that can exhibit low standard deviations linked to their CSR performance may appeal more to this growing demographic of investors, as it conveys greater predictability, less risk, and a commitment to ethical and sustainable practices.
### Caveats and Limitations
Stand-alone, the standard deviation provides limited insights into a company’s solutions for sustainability. Although a higher standard deviation might indicate inconsistency, this is not always an implication of failure to meet sustainable practices. A company may increase investments in CSR initiatives significantly during certain periods, leading to a high standard deviation. Consequently, it’s crucial to consider standard deviations in conjunction with other financial metrics and indicators to provide a holistic interpretation of a company’s CSR performance.
### Conclusion
Standard deviation thus offers a valuable, albeit limited and indirect, perspective on a company’s commitment to CSR. As a statistical method, it enriches the corporate sustainability assessment by capturing the volatility in financial performance linked to CSR initiatives. However, this method should be used in combination with other tools to affirm or counteract its insights.
## Standard Deviation vs Other Risk Measurement Metrics
### Comparison with Variance
While both standard deviation and variance measure dispersion in a dataset, it’s essential to note that standard deviation offers a clearer picture of the level of risk involved. The variance measures the average squared deviations from the mean, making its units challenging to interpret in terms of the original dataset. In contrast, standard deviation restores the measure to the original units of the data by taking the square root of the variance.
For example, suppose you are measuring volatility in daily changes in a stock’s price; variance could produce a figure in ‘dollars squared.’ It would then make more sense to the investor to discuss a standard deviation – in dollar terms – rather than variance.
### Comparison with Beta
While standard deviation is a measure of a security’s total risk, Beta measures systematic risk or market risk only. The standard deviation could be regarded as a more comprehensive measure as it takes into account both systematic risk (which Beta measures) and unsystematic risk (unique to each specific investment and can be reduced through diversification). Hence, standard deviation is a more reliable measure when considering investments in standalone assets, while Beta is more pertinent when constructing diversified portfolios.
### Comparison with Value-at-Risk (VaR)
VaR provides a measure of the maximum loss from a portfolio at a specific confidence level over a given period. Unlike standard deviation, which measures average risk, VaR outlines the worst-case scenario. Although VaR presents a clear picture of maximum potential loss, it is also silent on the size of the loss if it exceeds the VaR.
Additionally, VaR assumes normal distribution of returns, which might not always be the case in real-world financial markets. In contrast, standard deviation, being a measure of volatility, doesn’t assume any specific shape for the distribution of returns.
### Comparison with Conditional Value-at-Risk (CVaR)
While VaR describes the worst-case loss up to a certain confidence level, CVaR offers complementing insight by measuring the expected loss in the event the VaR threshold is breached. Standard deviation, on the other hand, doesn’t provide any insight on potential losses beyond a specific threshold.
So, when making risk assessments, financial practitioners might utilize a combination of standard deviation, VaR, and CVaR to capture a more holistic view of potential risks. This way, they can have a broad measure of volatility (standard deviation), a worst-case scenario (VaR), and understand what happens beyond that worst-case scenario (CVaR).
## Role of Standard Deviation in Regulatory Compliance
Regulators employ the concept of standard deviation in managing market risk strategies. The use of standard deviation helps to analyze, measure, and regulate the volatility pertaining to the returns of a financial instrument. It assists in deciphering the extent of probable variation in the yields over a certain period of time. Here, a higher standard deviation points towards a high degree of risk, and lower values denote stability.
Let us take a look at how the Basel Accords leverage this concept. The Basel regulations, enforced by the Bank of International Settlements (BIS), stress the importance of maintaining capital adequacies to address market risks. A key aspect of these accords, known as `Risk-Weighted Assets` (RWA), relies on the potential variation in asset values, which can be measured using standard deviation.
### Risk-Weighted Assets (RWA) and Standard Deviation
A risk weight represents the level of risk associated with a particular type of asset. Assets with higher risk are assigned a higher weightage. The formula for the calculation of Risk-Weighted Assets involves the consideration of standard deviation. As a result, the use of standard deviation allows for the consideration and calculation of risk-weighted assets, which directly influences the capital adequacy ratios under Basel regulations.
### Standard Deviation in Stress-Testing
Stress testing is another regulatory tool that uses standard deviation. Regulators carry out stress testing to assess the sustainability of a financial institution under extreme, but plausible, market conditions.
In conducting stress tests, regulators often examine the institution’s “Value at Risk” (VaR). Again, to calculate the VaR, standard deviation is taken into account. This enables them to capture the maximum potential loss over a set period of time under normal market conditions.
These are just a few examples of how standard deviation can contribute to regulatory compliance in financial markets, showcasing the concept’s significance in the broader finance sector. The next section will delve into the relevance of standard deviation in portfolio management.
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• This web page describes an activity within the Department of Mathematics at Ohio University, but is not an official university web page.
• If you have difficulty accessing these materials due to visual impairment, please email me at [email protected]; an alternative format may be available.
main calculus sage page
# Derivatives
You can plot a function and a secant line.
You can plot a function and secant lines tending toward a tangent line.
You can evaluate a derivative using its definition as a limit.
You can evaluate a derivative.
You can define a function, take its derivative, and plot both. Try changing to f(x)=(x-1)^2*sqrt(x) to see what happens. Then try f(x)=(x^5+x^3+2)/(8*x+1) ; change the interval to try to determine f'(0).
You can evaluate the derivative of a product with a manual product rule.
You can evaluate the derivative of a quotient with a manual quotient rule.
You can evaluate the derivative of a composition with a manual chain rule.
You can plot an implict curve $f(x,y)=0$. Try it for f(x,y)=2*y^3+y^2-y^5-x^4+2*x^3-x^2.
You can compute an implicit derivative.
You can evaluate a derivative involving exponentials, logs, hyperbolic functions, inverse trigonometric functions, ....
You can gather information to graph.
You can do Newton's method. Try setting p=0.5 and then p=3.0. Now try with f(x) = ((x-3/4)^(1/3))/(x^(1/3)) and p=0.1.
Martin J. Mohlenkamp
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A271834 a(n) = 2^n - Sum_{m=0..n} binomial(n/gcd(n,m), m/gcd(n,m)) = 2^n - A082906. 1
0, 0, 0, 4, 0, 42, 0, 116, 162, 730, 0, 2458, 0, 11494, 16890, 32628, 0, 180960, 0, 554994, 931476, 2800534, 0, 11005898, 6643750, 43946838, 44738892, 136580910, 0, 720879712, 0, 2147450740, 3250382916, 10923409738, 11517062060, 45683761528, 0, 172783692982 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 COMMENTS Compared to A082906, this sequence shows better the drop from 2^n upon replacing every binomial(n,m) in the Newton's expansion of (1+1)^n by the 'reduced' binomial(n/gcd(n,m), m/gcd(n,m)). For n > 1, a(n) is zero if and only if n is prime (no reduction, no drop). The ratio r(n) = a(n)/2^n is always smaller than 1 and presents considerable excursions. For composite n up to 5000, the minimum of 0.01471... occurs for n = 4489, and the maximum of 0.80849... occurs for n = 2310. This apparently large relative difference is actually surprisingly small: on log_2 scale it amounts to just about 5.78; a tiny fraction compared to the full scale, given by the values of n for the extrema. This insight suggests the following conjecture: there exists an average ratio r, defined as r = lim_{n->infinity} Sum_{m=1..n} r(m)/n. Its value appears to be approximately 0.3915+-0.0010, which can be interpreted as the average drop in a binomial value upon the 'reduction' of its arguments. LINKS Stanislav Sykora, Table of n, a(n) for n = 1..1000 Stanislav Sykora, Ratios A271834(n)/2^n for n=1..5000 FORMULA For prime p, a(p) = 0. For any n, a(n) < 2^n - n(n+1)/2. EXAMPLE Sum_{m=1..2500} r(m)/2500 = 0.391460... Sum_{m=2501..5000} r(m)/2500 = 0.391975... Sum_{m=1..5000} r(m)/5000 = 0.391718... MAPLE A271834:=n->2^n-add(binomial(n/gcd(n, m), m/gcd(n, m)), m=0..n): seq(A271834(n), n=1..50); # Wesley Ivan Hurt, Apr 19 2016 MATHEMATICA Table[2^n - Sum[Binomial[n/GCD[n, m], m/GCD[n, m]], {m, 0, n}], {n, 40}] (* Wesley Ivan Hurt, Apr 19 2016 *) PROG (PARI) bcg(n, m)=binomial(n/gcd(n, m), m/gcd(n, m)); a = vector(1000, n, 2^n-vecsum(vector(n+1, m, bcg(n, m-1)))) CROSSREFS Cf. A082905, A082906. Sequence in context: A271120 A174083 A123936 * A138546 A019217 A221757 Adjacent sequences: A271831 A271832 A271833 * A271835 A271836 A271837 KEYWORD nonn,easy AUTHOR Stanislav Sykora, Apr 19 2016 STATUS approved
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# Sequences I
In conventional math, an arithmetic sequence can be easily described by a simple algebraic expression when the common difference and the 0th term is known. This problem which was given during the 3rd International Vedic Mathematics Olympiad, Upper Primary group, illustrate how to determine the nth term of the sequence:
“A sequence starts, 2 8, 14, 20, 26, …. When it continues, what is the 100th number in the sequence?
A 582
B 588
C 594
D 596
E 602”
From the first five terms of the sequence, it is easy to see that the common difference, d, is 6. To determine the 0th term, we just need to subtract 6 from the first term, 2, to get – 4.
Thus the sequence can be described as 6n – 4 and the 100th term is 6(100) – 4 or 596. The intuitive solution is easier. After determining the common difference as 6, we just have to make the first term 6 by adding 4 to it. If we add 4 to each term of the sequence, we can see that all of them are multiples of 6. The 100th term is 100(6) or 600. Now if we deduct 4 from it, we will get 596, which is choice D.
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LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
subroutine zgeqr2 ( integer M, integer N, complex*16, dimension( lda, * ) A, integer LDA, complex*16, dimension( * ) TAU, complex*16, dimension( * ) WORK, integer INFO )
ZGEQR2 computes the QR factorization of a general rectangular matrix using an unblocked algorithm.
Purpose:
``` ZGEQR2 computes a QR factorization of a complex m by n matrix A:
A = Q * R.```
Parameters
[in] M ``` M is INTEGER The number of rows of the matrix A. M >= 0.``` [in] N ``` N is INTEGER The number of columns of the matrix A. N >= 0.``` [in,out] A ``` A is COMPLEX*16 array, dimension (LDA,N) On entry, the m by n matrix A. On exit, the elements on and above the diagonal of the array contain the min(m,n) by n upper trapezoidal matrix R (R is upper triangular if m >= n); the elements below the diagonal, with the array TAU, represent the unitary matrix Q as a product of elementary reflectors (see Further Details).``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,M).``` [out] TAU ``` TAU is COMPLEX*16 array, dimension (min(M,N)) The scalar factors of the elementary reflectors (see Further Details).``` [out] WORK ` WORK is COMPLEX*16 array, dimension (N)` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```
Date
September 2012
Further Details:
``` The matrix Q is represented as a product of elementary reflectors
Q = H(1) H(2) . . . H(k), where k = min(m,n).
Each H(i) has the form
H(i) = I - tau * v * v**H
where tau is a complex scalar, and v is a complex vector with
v(1:i-1) = 0 and v(i) = 1; v(i+1:m) is stored on exit in A(i+1:m,i),
and tau in TAU(i).```
Definition at line 123 of file zgeqr2.f.
123 *
124 * -- LAPACK computational routine (version 3.4.2) --
125 * -- LAPACK is a software package provided by Univ. of Tennessee, --
126 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
127 * September 2012
128 *
129 * .. Scalar Arguments ..
130 INTEGER info, lda, m, n
131 * ..
132 * .. Array Arguments ..
133 COMPLEX*16 a( lda, * ), tau( * ), work( * )
134 * ..
135 *
136 * =====================================================================
137 *
138 * .. Parameters ..
139 COMPLEX*16 one
140 parameter ( one = ( 1.0d+0, 0.0d+0 ) )
141 * ..
142 * .. Local Scalars ..
143 INTEGER i, k
144 COMPLEX*16 alpha
145 * ..
146 * .. External Subroutines ..
147 EXTERNAL xerbla, zlarf, zlarfg
148 * ..
149 * .. Intrinsic Functions ..
150 INTRINSIC dconjg, max, min
151 * ..
152 * .. Executable Statements ..
153 *
154 * Test the input arguments
155 *
156 info = 0
157 IF( m.LT.0 ) THEN
158 info = -1
159 ELSE IF( n.LT.0 ) THEN
160 info = -2
161 ELSE IF( lda.LT.max( 1, m ) ) THEN
162 info = -4
163 END IF
164 IF( info.NE.0 ) THEN
165 CALL xerbla( 'ZGEQR2', -info )
166 RETURN
167 END IF
168 *
169 k = min( m, n )
170 *
171 DO 10 i = 1, k
172 *
173 * Generate elementary reflector H(i) to annihilate A(i+1:m,i)
174 *
175 CALL zlarfg( m-i+1, a( i, i ), a( min( i+1, m ), i ), 1,
176 \$ tau( i ) )
177 IF( i.LT.n ) THEN
178 *
179 * Apply H(i)**H to A(i:m,i+1:n) from the left
180 *
181 alpha = a( i, i )
182 a( i, i ) = one
183 CALL zlarf( 'Left', m-i+1, n-i, a( i, i ), 1,
184 \$ dconjg( tau( i ) ), a( i, i+1 ), lda, work )
185 a( i, i ) = alpha
186 END IF
187 10 CONTINUE
188 RETURN
189 *
190 * End of ZGEQR2
191 *
subroutine zlarfg(N, ALPHA, X, INCX, TAU)
ZLARFG generates an elementary reflector (Householder matrix).
Definition: zlarfg.f:108
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine zlarf(SIDE, M, N, V, INCV, TAU, C, LDC, WORK)
ZLARF applies an elementary reflector to a general rectangular matrix.
Definition: zlarf.f:130
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Q:
# How do I calculate capital gains yield?
A:
The capital gains yield of a stock can be calculated by dividing the change in price of the stock after the first period by the original price. Investopedia explains that the formula for this is (P1 - P0) / P0, where P1 equals the original price paid and P0 equals the price after the first period.
## Keep Learning
Capital gains yield is the appreciation of the stock over a certain period (which may be annual, monthly or quarterly), whereas dividends are determined by the company and paid out to shareholders during a certain period. For example, if Company X's stocks are purchased at \$10 per share and they increase to \$100 per share during the quarter, the capital gains yield formula would be 100 - 10 / 10 = 9, or 900 percent.
When the capital gains yield occurs over multiple periods, each period cannot be summed to generate a total capital gains yield. Instead, the investor must use the holding period return formula. This formula is HPR = [(1 + r1) * (1 + r2) * ... (1 + rx)] - 1, where r is the return per period and rx is the number of periods.
When investing in a stock, it is important to calculate the holding period return rate rather than simply looking at the capital gains yield or the dividend return rate, because it is possible that stocks with lower dividend returns held over shorter periods of time may provide a greater percentage of return on the investment.
Sources:
## Related Questions
• A:
A change in net working capital is a result of a change in either the current assets or current liabilities without a similar change in the other figure. The change occurs because the net working capital is calculated as current assets minus current liabilities.
Filed Under:
• A:
In order to calculate the market price per share, you must know the price per share on a given date as well as the number of outstanding shares of stock. You also have to find out if the stock pays dividends. Put these values in a formula to determine the market price per share.
Filed Under:
• A:
Calculate a stock's dividend yield by dividing the annual dividend per share by the price per share for that stock, explains About.com. If company XYZ pays a yearly dividend of \$2 per share while the stock trades at \$100 per share, the dividend yield is 2 percent, or 2 divided by 100.
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This article describes the formula syntax and usage of the CUMPRINC function in Microsoft Excel.
## Description
Returns the cumulative principal paid on a loan between start_period and end_period.
## Syntax
CUMPRINC(rate, nper, pv, start_period, end_period, type)
The CUMPRINC function syntax has the following arguments:
• Rate Required. The interest rate.
• Nper Required. The total number of payment periods.
• Pv Required. The present value.
• Start_period Required. The first period in the calculation. Payment periods are numbered beginning with 1.
• End_period Required. The last period in the calculation.
• Type Required. The timing of the payment.
Type Timing 0 (zero) Payment at the end of the period 1 Payment at the beginning of the period
## Remarks
• Make sure that you are consistent about the units you use for specifying rate and nper. If you make monthly payments on a four-year loan at an annual interest rate of 12 percent, use 12%/12 for rate and 4*12 for nper. If you make annual payments on the same loan, use 12% for rate and 4 for nper.
• If rate ≤ 0, nper ≤ 0, or pv ≤ 0, CUMPRINC returns the #NUM! error value.
• If start_period < 1, end_period < 1, or start_period > end_period, CUMPRINC returns the #NUM! error value.
• If type is any number other than 0 or 1, CUMPRINC returns the #NUM! error value.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data Description 0.09 Interest rate per annum 30 Term in years 125000 Present value Formula Description Result =CUMPRINC(A2/12,A3*12,A4,13,24,0) The total principal paid in the second year of payments, periods 13 through 24 -934.1071234 =CUMPRINC(A2/12,A3*12,A4,1,1,0) The principal paid in a single payment in the first month (-68.27827) -68.27827118
Notes:
• You divide the interest rate by 12 to get a monthly rate, and you multiply the years the money is paid out by 12 to get the number of payments.
• In Excel Web App, to view the result in its proper format, select the cell, and then on the Home tab, in the Number group, click the arrow next to Number Format, and click General.
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# 881: Probability
Probability Title text: My normal approach is useless here, too.
## Explanation
Cueball and Megan are sitting on a hospital bed, reading a piece of paper with the statistics for breast cancer survival. It looks like Megan has just been diagnosed with breast cancer. The thick line represents the survival rate distribution (probability to be alive after X years, unconditioned): 81% are alive at 5 years, while 77% survive to 10 years. The dashed line represents the hazard function (the negative derivative of the thick line divided by the value of the thick line at each point, i.e. how fast the thick line falls with respect to the current value, or the risk of failing/dying at time t+Δt after having survived until time t as Δt approches zero), which is the rate between the density of the failure distribution and the survival function.
Randall wrote this comic after his fiancee was diagnosed with breast cancer. Two months after posting this strip, he posted this blog post explaining the cancer strips.
The title text could be a reference to the comic 55: Useless. Cueball's (and Randall's) normal approach — math — isn't much help in dealing with this emotional situation, either. The phrase "normal approach" may also in this case be a pun on the Normal distribution, which is another probability distribution that is commonly used in statistics. Interpreted this way, the title text states that the "Normal distribution" is not used as a hazard function.
## Transcript
[A plot of percent vs. years, with a solid and a dashed line. The solid line starts at 100%, and drops constantly. The dashed line starts around 85%, rises to 95% after 5 years, then drops.]
[A simple table.]
5 years 81% 10 years 77%
[Cueball and Megan are sitting on a bench, next to an Intravenous drip hanging from a rack. Cueball is holding a paper.]
Cueball: You know, probability used to be my favorite branch of math
Cueball: Because it had so many real-life applications.
[They embrace, faces together.]
# Discussion
I've been through this, even though it was a little over a year ago, this strip brings me back to how I tried to rationalise the probabilities to deal with the news, and the only thing I could think of was "I want a better number, god, noodle-monster, anybody, please, give me a better number".
John 60.225.31.6 00:40, 28 September 2013 (UTC)
Can anyone please provide an update to this page on Randall's fiance's health? Apologies if this is common knowledge. I agree with John's reaction. I wanted a better number for my wife back in the spring of 2008, and got it. She survived 3 years instead of the (then) predicted average of six months for inflammatory breast cancer. We could have been just an outlier on the low probability end of the curve, but I like to think the medical community is continually improving their curves, and I am very grateful for the extra time. She passed away four days after this strip was posted - which explains why I haven't seen this strip until now.
108.162.216.228 21:26, 28 November 2013 (UTC)Grant
So, how is this incomplete? Can we remove the incompleteness mark? --173.245.53.196 13:02, 19 January 2014 (UTC)
I think the linking to the blog post is more sensitive than explaining it here. Remove tag? --141.101.98.207 14:19, 19 January 2014 (UTC)
What is the dashed line? It kind of looks like it might be the derivative of the solid line.108.162.214.53 00:39, 20 January 2014 (UTC)
I think I explained the dashed line, it's a hazard function or at least it would be a plausible hazard function for that kind of survival function. Feel free to improve the formatting or remove the incomplete tag. --Artod (talk) 07:06, 20 January 2014 (UTC)
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# The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm.
Question:
The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm. The difference of their total surface areas is
(a) 38 cm2
(b) 58 cm2
(c) 78 cm2
(d) 88 cm2
Solution:
(d) 88 cm2
Suppose that the radii of the spheres are cm and (7 − r) cm. Then we have:
$\frac{\frac{4}{3} \pi(7-r)^{3}}{\frac{4}{3} \pi r^{3}}=\frac{64}{27}$
$\Rightarrow \frac{(7-\tau)}{r}=\frac{4}{3}$
$\Rightarrow 21-3 r=4 r$
$\Rightarrow 21=7 r$
$\Rightarrow r=3 \mathrm{~cm}$
Now, the radii of the two spheres are 3 cm and 4 cm.
$\therefore$ Required difference $=4 \times \frac{22}{7} \times 4^{2}-4 \times \frac{22}{7} \times 3^{2}$
$\therefore$ Required difference $=4 \times \frac{22}{7} \times 4^{2}-4 \times \frac{22}{7} \times 3^{2}$
$=4 \times \frac{22}{7} \times(16-9)$
$=4 \times \frac{22}{7} \times 7$
$=88 \mathrm{~cm}^{2}$
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# 9.4 Applications of statics, including problem-solving strategies (Page 2/3)
Page 2 / 3
Similar observations can be made using a meter stick held at different locations along its length.
If the pole vaulter holds the pole as shown in [link] , the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If ${F}_{L}={F}_{R}$ , then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces ${F}_{L}$ and ${F}_{R}$ is straightforward, as the next example shows.
If the pole vaulter holds the pole from near the end of the pole ( [link] ), the direction of the force applied by the right hand of the vaulter reverses its direction.
## What force is needed to support a weight held near its cg?
For the situation shown in [link] , calculate: (a) ${F}_{R}$ , the force exerted by the right hand, and (b) ${F}_{L}$ , the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.
Strategy
[link] includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium $\text{(net}\phantom{\rule{0.25em}{0ex}}F=0$ ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium $\text{(net}\phantom{\rule{0.25em}{0ex}}\tau =0\text{)}$ if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.
Solution for (a)
There are now only two nonzero torques, those from the gravitational force ( ${\tau }_{\text{w}}$ ) and from the push or pull of the right hand ( ${\tau }_{R}$ ). Stating the second condition in terms of clockwise and counterclockwise torques,
$\text{net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{cw}}=\text{–net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{ccw}}\text{.}$
or the algebraic sum of the torques is zero.
Here this is
${\tau }_{R}={\mathrm{–\tau }}_{\text{w}}$
since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, $\tau =\text{rF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , noting that $\theta =90º$ , and substituting known values, we obtain
$\left(0\text{.}\text{900 m}\right)\left({F}_{R}\right)=\left(0\text{.600 m}\right)\left(\mathit{mg}\right)\text{.}$
Thus,
$\begin{array}{lll}{F}_{R}& =& \left(0.667\right)\left(\text{5.00 kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\\ & =& \text{32.7 N.}\end{array}$
Solution for (b)
The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:
${F}_{L}+{F}_{R}–\text{mg}=0$
From this we can conclude:
${F}_{L}+{F}_{R}=w=\text{mg}$
Solving for ${F}_{L}$ , we obtain
$\begin{array}{lll}{F}_{L}& =& \mathit{mg}-{F}_{R}\\ & =& \mathit{mg}-\text{32}\text{.}7 N\\ & =& \left(\text{5.00 kg}\right)\left(\text{9.80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)-\text{32.7 N}\\ & =& \text{16.3 N}\end{array}$
Discussion
${F}_{L}$ is seen to be exactly half of ${F}_{R}$ , as we might have guessed, since ${F}_{L}$ is applied twice as far from the cg as ${F}_{R}$ .
If the pole vaulter holds the pole as he might at the start of a run, shown in [link] , the forces change again. Both are considerably greater, and one force reverses direction.
## Take-home experiment
This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity!
## Phet explorations: balancing act
Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game.
## Summary
• Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply.
## Conceptual questions
When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person’s neck vertebrae.
## Problems&Exercises
To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?
In [link] , the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in [link] , show that the second condition for equilibrium $\text{(net}\phantom{\rule{0.25em}{0ex}}{\tau }_{}=\text{0)}$ is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above.
if the mass of a trolley is 0.1kg. calculate the weight of plasticine that is needed to compensate friction. (take g=10m/s and u=0.2)
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i don't know
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They are interchangeable.
Shii
Distance is scalar, displacement is vector because it must involve a direction as well as a magnitude. distance is the measurement of where you are and where you were displacement is a measurement of the change in position
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Thanks a lot
Usman
I'm beginner in physics so I can't reason why v=u+at change to v2=u2+2as and vice versa
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what is kinematics
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kinematics is study of motion without considering the causes of the motion
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The study of motion without considering the cause 0f it
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why electrons close to the nucleus have less energy and why do electrons far from the nucleus have more energy
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I have not try that experiment but I think it will magnet....
Hey Rev. it will
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I do think so, it will
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yes it will
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If a magnet is in a pool of water, would it be able to have a magnetic field?.
yes Stella it would
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formula for electric current
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I=q/t
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Current is the flow of electric charge per unit time.
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What are semi conductors
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materials that allows charge to flow at varying conditions, temperature for instance.
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wao so awesome
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it is branch of science that deal with interaction matter and energy is called physics . and physics is based in experiential observation and quentative measurement.
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to briefly understand the concept of physics start with history and a brief history of time by Stephen hawkings is what made me have interest in physics
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physics is a branch of science which deals with the study of matter, in relation to energy.
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physics is a natural science that involve the study of matter and it's motion through space and time, along with related concept such as energy and force
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Physics is the science of natural things. for instance, take classical laws which describe the principles of working of the macro realm and then take the quantum laws which describe the quantum realm. It relates everything in this universe –e. g when you see anything, actually photons penetrate.
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what is the purpose of finding electron temperature and electron number density of an element in LIBS
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Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/13) | Q#10
Hits: 585
Question
The function is defined by for .
i. Find and and hence verify that the function has a minimum value at .
The points and lie on the curve , as shown in the diagram.
i. Find the distance AB.
ii. Find, showing all necessary working, the area of the shaded region.
Solution
i.
We are given;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
A stationary point on the curve is the point where gradient of the curve is equal to zero;
If the given function has a minimum (stationary) value at then at .
We have found above that for given function ;
At ;
Therefore, function has a stationary value at .
Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2nd derivative of the curve.
We substitute of the stationary point in the expression of 2nd derivative of the curve and evaluate it;
If or then stationary point (or its value) is minimum.
If or then stationary point (or its value) is maximum.
We have found above that for given function ;
At ;
Since , function has a minimum value at .
ii.
Expression to find distance between two given points and is:
Therefore for points and ;
iii.
It is evident from the diagram that;
To find the area of region under the curve , we need to integrate the curve from point to along x-axis.
We are given equation of the curve as follows;
However, we do not have equation of line but using points and we can write equation of line AB.
Two-Point form of the equation of the line is;
Therefore;
Now we can find area of shaded region.
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
Rule for integration of is:
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You are Here: Home >< A-levels
# S2 question 12, exercise 7D. watch
1. So, I did the question and I came to a similar answer to the book and the same conclusion however our approaches were different and the answers were not completely parallel, and this is due to the angle we took the question from.
Heres the question: At one stage of a water treatment process the number of particles of foreign matter per litre present in the water has a poisson distribution with mean 10. The water then enters a filtration bed which should extract 75% of foreign matter. The manager of the treatment works orders a study into the effectiveness of this filtration bed. Twenty samples, each 1 litre, are taken from the water and 64 particles of foreign matter are found.. Using a suitable approximation test, at the 5% level of significance, whether or not there is evidence that the filter bed is failing to work properly.
In the book, they base the normal distribution on a derived poisson. They say the average amount of particles passing through the filtration bed is 50 and get this by finding the average amount of particles in 20 litres (20*10) then multiplying it by 0.25 (the proportion that is said to get through). They then approximate using a normal distribution defined by X-N(50,50) and then get an answer of 0.0282.
Now what I did, was find the average of particles in 20 litres (200), and then model the entire thing as a binomial distribution using the 0.75 as a probability of a particle being removed. Defined X-B(200,0.75).
I then approximated using a normal distribution and that's where my method and the books method diverged in terms of values. My derived normal distribution was X-N(50,37.5). As you can see, the variances differ. Therefore, this brought me to a slightly smaller value of Z(-2.20) and my final probability was 0.0139 (I flipped it around and said 136 particles did not get filtered then found P(X<136.5). Both of our resultant probabilities were smaller than the specified 0.05 significance level therefore we both arrived at the same conclusion, that the filtration bed was not working as well as he thought.
Would I get penalised for taking a difference approach about the question and not getting the same answer? As far as I'm concerned my method is as legit as theirs.
What do you guys think?
Why not try posting in a specific subject forum- you might have more luck there.
Here's a link to our subject forum which should help get you more responses.
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# Connected Particles Question
1. Apr 17, 2004
### mcintyre_ie
Hey,
I’d appreciate some help with this question. Here’s a diagram:
Diagram
And the question:
A moveable pulley of mass 2Kg is suspended on a light inextensible string between two fixed pulleys as shown and masses of 6kg and 3kg are asttached to the ends of the string. If the system is released from rest:
(I) Show in a diagram the forces acting on each of the masses.
(II) Find the acceleration of the moveable pulley and the tension in the string.
(III) If initially the moveable pulley had been replaced by another of mass m, find m, given that the moveable pulley remains at rest while the other two masses are in motion
I’m a bit confused about the accelerations of each particle and the pulley. From what I understand, the 6kg particle will move downwards, and the pulley and 3kg particle will move upwards. I’m assuming that some particle (probably the pulley?) will have a different acceleration to the others, maybe twice that of the others? I’m just lost as to what the accelerations are.
(I)Should be fine if I can find the accelerations.
(II)Should be ok too, once I have the accelerations and then use F=MA.
(III)Seems complicated, and I’m feeling I’ll probably have difficulty.
So, any advice on the accelerations and part (III) would be appreciated.
#### Attached Files:
• ###### Diagram010.jpg
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2. Apr 18, 2004
### Staff: Mentor
First label all the forces: the tension in the string plus the weights of the masses.
For simplicity, I would assume that the masses at each end accelerate downward: a1 and a2. (Note: Make an educated guess as to how it will move--if wrong, the acceleration will be negative.) The acceleration of the middle pulley is related to that of the masses by: a = (a1 + a2)/2 upward. Make sure you understand this. (Think what happens if both masses move down by a given distance--what happens to the middle pulley?)
3. Apr 18, 2004
### mcintyre_ie
I was thinking that the 6kg mass would be the only mass to move downwards, but both the other mass and the pulley would move upwards? For some reason I was thinking that the pulley would have an acceleration of half of that of each of the masses.
Last edited: Apr 18, 2004
4. Apr 18, 2004
### Staff: Mentor
No need to guess. Figure it out! To start, assume some direction for the acceleration of the the two masses. This will define your sign convention for applying F=ma. If you chose correctly, the accelerations will come out positive. Try it and see.
I'm not sure what you mean. The motion of the pulley is related to the motion of the masses. See my previous post. To see the relationship, use some examples: if the left mass drops by a distance X1, and the right one rises by X2, then how must the pulley move? Don't forget that the rope is doubled over the middle pulley. (Note that I used your assumption of how the masses move.)
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# How Is Add On Interest Calculated?
Add-on interest is a method of calculating the interest to be paid on a loan by combining the total principal amount borrowed and the total interest due into a single figure, then multiplying that figure by the number of years to repayment.
The total is then divided by the number of monthly payments to be made.
## What is meant by add on interest?
Add on interest is a method of charging interest. Interest payable is determined at the beginning of a loan. In add on method interest, interest amount is added on the total amount borrowed and added on to the principal of a debt. Then each payment is deducted from the total amount.
## What is add on rate per month?
Loans with add-on interest is paid in equal installments every month, and principal and interest payments are also constant monthly. The principal repayment portion comes from dividing the principal by the number of payment periods; in our example, it’s 30,000 / 12 months = 2,500 pesos per month.
## What’s the difference between simple interest and add on interest?
Simple interest is based on the principal amount of a loan or deposit, while compound interest is based on the principal amount and the interest that accumulates on it in every period. Since simple interest is calculated only on the principal amount of a loan or deposit, it’s easier to determine than compound interest.
## What is an add on rate or margin?
An ARM margin is a fixed percentage rate that is added to an indexed rate to determine the fully indexed interest rate of an adjustable rate mortgage (ARM). Adjustable rate mortgages are one of the most common variable rate credit products offered in the primary lending market.
## What is the add on rate?
Typically, most banks and financing companies will quote you an add-on interest rate, which is the gross interest rate per year multiplied by the number of years of the loan. This add-on interest is added to your principal loan amount and divided by the number of repayments, which is equal to your amortization.
## How do you add up interest?
Divide your interest rate by the number of payments you’ll make in the year (interest rates are expressed annually). So, for example, if you’re making monthly payments, divide by 12. 2. Multiply it by the balance of your loan, which for the first payment, will be your whole principal amount.
## What is the rule of 78s prepayment penalty?
If the borrower pays off the loan early, this method maximizes the amount paid (interest paid) by applying funds to interest before principal. In other words, in comparison to a simple interest loan, a rule of 78s loan will charge more interest if the loan is paid early.
## How do you calculate monthly interest rate?
• Convert the annual rate from percentage to decimal format (by dividing by 100)
• 10/100 = 0.1 annually.
• Divide the annual rate by 12.
• 0.10/12 = .0083.
• Calculate the monthly interest on \$100.
• 0.0083 x \$100 = \$0.83.
• Convert the monthly rate in decimal format back to a percentage (by multiplying by 100)
## What is simple interest?
Simple interest is a quick and easy method of calculating the interest charge on a loan. Simple interest is determined by multiplying the daily interest rate by the principal by the number of days that elapse between payments.
## Which interest is better simple or compound?
When it comes to investing, compound interest is better since it allows funds to grow at a faster rate than they would in an account with a simple interest rate.
## What is simple interest example?
Examples : 1) Ariel takes a loan of \$8,000 to buy a used truck at the rate of 9 % simple Interest.Calculate the annual interest to be paid for the loan amount. Solution: From the details given in the problem Principle = P = \$8,000 and R = 9% or 0.09 expressed as a decimal.
## What is simple daily interest?
Interest on a daily simple interest loan is calculated by using the daily simple interest method. This means that interest accrues on a daily basis on the amount of the loan (current outstanding principal balance) from the date the interest charges begin until you repay the loan.
## What is margin in lending?
Margin lending explained. A margin loan is a type of investment loan that lets you borrow money to invest in shares, managed funds and other approved financial products. It’s important to weigh up both the benefits and the risks when thinking about investing with a margin loan.
## What is the difference between the margin and the index?
A mortgage margin is the difference between the index and the interest rate charged for a particular loan. The margin is a fixed percentage point that is predetermined by the lender and added to the index to compute the interest rate. A lender’s margin remains fixed for the entire term of the loan.
## What is the prime lending rate today?
Prime rate, federal funds rate, COFI
This weekMonth ago
WSJ Prime Rate5.505.50
Federal Discount Rate3.003.00
Fed Funds Rate (Current target rate 2.25-2.50)2.502.50
11th District Cost of Funds1.141.14
## What is the formula for compound interest?
Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one.
## What is a simple loan?
Simple interest applies mostly to short-term loans, such as personal loans. A simple-interest mortgage charges daily interest instead of monthly interest. When the mortgage payment is made, it is first applied to the interest owed. Any money that’s left over is applied to the principal.
## What does effective interest rate mean?
The effective annual interest rate is the interest rate that is actually earned or paid on an investment, loan or other financial product due to the result of compounding over a given time period. It is also called the effective interest rate, the effective rate or the annual equivalent rate.
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# • A company makes two kinds of leather belts. Belt A is high quality belt and belt B is of lower quality.
Each belt of type A requires twice as much time as a belt of type B and if all the belts are of type B, the company could make 10,000 belts per day. The supply of leather is sufficient for only 8,000 belts per day(both A and B combined). Belt A requires a fancy buckle and only 4000 per day are available. There are only 7000 buckles a day available for belt B. What should be the daily production of each type of belt if the respective profits of belt A and B are Rs. 40 and Rs. 30.
000 (Availability of buckles for belt B) x.Decision variables: Let x and y be the quantity of belt A and belt B be produced respectively. Objective function: Maximize Z = 40x + 30y Subject to constraints: 2x+y ≤ 10000 (Total availability of time) x + y ≤ 8000 (Total availability of leather) x ≤ 4000 (Availability of buckles for belt A) y ≤ 7. y ≥ 0 (Non-negativity constraint) to .
• The optimum solution to this problem is x = 2000 y = 6000 which means that 2000 units of belt A 6000 units of belt B should be produced. . 2. This will give a profit of Rs.60.000.
• A dealer wishes to purchase a number of boxes of center fresh and mentos.25 times of space than center fruit and he has space utmost for 24 units. Assuming that he can sell all the items that he can buy. His expectation is that he can sell a box of center fresh at a profit of ₹ 5 per box and center fruit at a profit of ₹ 7 per box. He has only ₹ 3000 to invest. how should he invest his money in order to maximize his profit? . Market demand for center fresh and center fruit is 10 units and 20 per week respectively. Center fresh requires 1. A box of center fresh costs ₹ 150 and a box of center fruit costs ₹ 100.
y ≥ 0 (Non-negativity constraint) . Objective function: Maximize Z = 5x + 7y Subject to constraints: 150x + 100y ≤ 3000 (Total availability of amount) 1.Decision variables: Let x and y be the quantity of center fresh and center fruit to be purchased respectively.25x + y ≤ 24 (Total availability of space) x ≤ 10 (Market demand for center fresh) y ≤ 20 (Market demand for center fruit) x.
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# hw4ANSWERS - Neal Whittington(whittin3 Tuesdays at 4:00PM...
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Neal Whittington (whittin3) Tuesdays, at 4:00PM AD9 CS 173: Discrete Structures, Spring 2010 Homework 4 This homework contains 5 problems worth a total of 50 points. It is due on Friday, 19 February at 4pm. 1. Set Operations [16 points] Suppose you were given the following sets: A = { Piano , { Violin , Viola , Cello } , Guitar } B = {{ Flute, Piccolo } , Cymbals } C = { Piano , Flute } D = {{ Violin, Viola, Cello } , { Flute, Piccolo }} List the elements of the set for the following expressions: (a) A D { Piano , { Violin , Viola , Cello } , Guitar , { Flute, Piccolo }} (b) B C (c) A - ( B - C ) { Piano, { Violin, Viola, Cello }} (d) A P ( B C ) (e) ( B D ) × C {{ Flute, Piccolo } × { Piano, Flute } = {{ Flute, Piano } , { Flute, Flute } , { Piccolo, Piano } , { Piccolo, Flute }} (f) | P ( B D ) | {{ Flute } , { Piccolo } , { ∅ } , { Flute, Piccolo }} (g) { X P ( A ) : | X | is not prime } (h) { X ( P ( A ) P ( B )) : | X | ≡ 3 (mod 2) } 1
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2. Euclidean algorithm [4 points] x y r 837 2013 837 2013 837 341 341 155 31 155 31 0 31 0 gcd (837 , 2015) is 31. 3. Pseudocode [10 points] (a) Trace the execution of func(a, b) . a b m p return 2 5 func(2, 2) = 4 func(2, 1) = 2 32 2 2 func(2, 1) = 2 func(2, 0) = 1 4 2 1 2 2 0 1 32 = 2 5
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# Basic projectile motion problem
by nick727kcin
Tags: basic, motion, projectile
P: 160 ... You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 63.0 m away, making a 3.00 degree angle with the ground. how fast was the arrow shot? look two posts down....
P: 160 using another equation.... d= vit +.5at^2 63= vi(.616)+.5(-187)(.616)^2 vi= 678.23m/s which is also wrong... in case a wasnt supposed to be negative: 63=vi(.616)+.5(187)(.616^2) vi= -473.69 m/s which is also wrong
PF Gold P: 867 I'm not sure exactly what you did but for one thing, there is no acceleration in the x direction, only downwards cause of energy. So the Vox = Vix. Cause of that you know that the x component of the original speed times the time equals the distance (63). Also, the y component of the original speed Viy goes with the equation: H = Viy*t - 0.5g*t^2. The H you have (dy, but I got 3.3) so now you have two equations with two unknowns (t and Vi).
P: 160 Basic projectile motion problem ok so this is the real time: time=.8208s dx= 3.30 m dx= vt v= 76.85 m/s but this is wrong. ahhh this is killll
Sci Advisor HW Helper PF Gold P: 12,016 First, be absolutely clear on what are your principal unknowns here! Those are a) the initial velocity V in the x-direction and b) The height H from which the archer shot the arrow parallell to the ground (i.e, the initial vertical velocity is, indeed 0, as you thought). c) The time t it took the arrow to hit the ground. Let's set up a few equations relating V,t,H: Horizontal distance traveled: $$63=Vt$$ Vertical distance traveled: $$H=\frac{gt^{2}}{2}$$ Ratio of velocities at time of impact: $$\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V$$ The last equation expresses that the arrow at the time of impact follows is parallell to the tangent line of the parabolic arc at that point. You only need the first and third equation to answer your question.
P: 160
Quote by arildno First, be absolutely clear on what are your principal unknowns here! Those are a) the initial velocity V in the x-direction and b) The height H from which the archer shot the arrow parallell to the ground (i.e, the initial vertical velocity is, indeed 0, as you thought). c) The time t it took the arrow to hit the ground. Let's set up a few equations relating V,t,H: Horizontal distance traveled: $$63=Vt$$ Vertical distance traveled: $$H=\frac{gt^{2}}{2}$$ Ratio of velocities at time of impact: $$\frac{V_{y}}{V_{x}}=-\tan(3), V_{y}=-gt, V_{x}=V$$ The last equation expresses that the arrow at the time of impact follows is parallell to the tangent line of the parabolic arc at that point. You only need the first and third equation to answer your question.
t=.8209s
Vy= -8.0444m/s
final result:
Vy/[-tan(3)]=Vx
Vx=[-gt]/[tan(3)]
vx=153.5 m/s
its still saying its wrong, but thanks alot man
Sci Advisor HW Helper PF Gold P: 12,016 Eeh, not too sure what you've done here! From the 3rd equation, we get: $$t=V\frac{\tan(3)}{g}$$ and hence, from the first: $$V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60 }\sqrt{3}$$
P: 160
Quote by arildno Eeh, not too sure what you've done here! From the 3rd equation, we get: $$t=V\frac{\tan(3)}{g}$$ and hence, from the first: $$V=\sqrt{\frac{63g}{\tan(3)}}\approx\sqrt{\frac{60*63*g}{\pi}}\approx{60 }\sqrt{3}$$
ooo i get it. you substituted vy=-gt into the third equation. thanks so much man:
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### what percent that 350 is 75 ?
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# Mean Simplifying Algebraic Expressions Worksheet Tes
Math Worksheets. September 12th , 2020.
My Algebra 1 students just finished our unit on systems of
### The questions will provide you with a variety of algebraic expressions.
Simplifying algebraic expressions worksheet tes. The quiz is a collection of math problems. They have been taught to a variety of abilities, so find the ppt that suits your class best. Algebraic expressions can be simplified by collecting like terms or expanding (multiplying) or factorising (dividing by the highest common factor).
Collecting like terms with 1 variable (l4) section b: Simplifying expressions | finding perimeter of quadrilaterals. These algebraic expressions worksheets will create algebraic statements for the student to simplify.
Combine like terms and use the order of operations to simplify algebraic. A variety of powerpoints, worksheets and activities i have made over the past year to teach simplifying expressions. Algebraic expressions can be added and subtracted by collecting like terms, but expressions can also be multiplied and divided.
Expressions in maths year 9. Original resource here by mr pullin. Education resources, designed specifically with parents in mind
7 questions | by dtullo | last updated: In most cases the expressions have parentheses. This page starts off with some missing numbers worksheets for younger students.
Jun 21, 2019 | total attempts: These algebraic expressions worksheets are a good resource for students in the 5th grade through the 8th grade. This quiz and attached worksheet will gauge your understanding of simplifying algebraic expressions with negative signs.
The worksheets can be made either as pdf or html files (the latter are editable in a word processor). Use this brilliant collection of algebraic expressions worksheets to practise key skills in simplifying algebraic expressions! You may select from 2, 3, or 4 terms with addition, subtraction, and multiplication.
These exercises help students get accustomed to reducing expressions that have basic math operations. Example 1 simplify $$a \times a$$. Keystage 3 interactive worksheets to help your child understand algebra:
Algebraic index notation laws of indices laws of indices laws of indices laws of indices (no answers) laws of indices 2 laws of indices dominoes lesson activities keyboard_arrow_up Simplifying algebraic expressions, expanding brackets, solving linear equations, applications. Use these algebraic expressions worksheets to introduce or recap on simplifying algebraic expressions using powers and the four operations. there are 4 separate 'how can i simplifying algebraic expressions worksheets' that cover different skills, including: collecting terms multiplyingdividing powersalso included is a mixed exercise on simplifying algebraic expressions.
Sequential easy first hard first. You will then need to simplify the given expression. See more ideas about algebraic expressions, simplifying algebraic expressions, simplifying expressions.
Whenever you see a number or variable with a negative exponent, move it to the opposite. About this quiz & worksheet. The assemblage of printable algebra worksheets encompasses topics like translating phrases, evaluating and simplifying algebraic expressions, solving equations, graphing linear and quadratic equations, comprehending linear and quadratic functions, inequalities.
Collecting like terms with more than 1 variable (l5) section c: Add the side lengths, simplify the algebraic expressions and express the perimeter in expression. Algebra is a branch of math in which letters and symbols are used to represent numbers and quantities in formulas and equations.
Simplify numerical expressions involving multiple basic math operations. This gives them a hint about which operation should be done first. A trivia quiz on simplifying algebraic expressions.
On this page, you will find algebra worksheets mostly for middle school students on algebra topics such as algebraic expressions, equations and graphing functions.
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# Maxwell’s Equations
There are four fundamental equations of electromagnetism known as Maxwell’s equations
which may be written in differential form as
1. div D = p
2. div B = 0
3. curl E = – dB /dt
4. curlH =J + -dD / dt
In above equations the notations have the following meanings:
J =current density in Amp/met3
D =electric displacement vector in Coulomb/met2
p = free charge density in Coul / met3
B =magnetic induction in Weber/met2
E = electric field intensity in Volt/met or Newton/Coul
H =magnetic field intensity in Amp/met-tum.
7.16.1 Derivation of Maxwell’s Equations
1. Derivation of First Equation
div D = ∆.D = p
Proof: “The maxwell first equation .is nothing but the differential form of Gauss law of electrostatics.”
Let us consider a surface S bounding a volume V in a dielectric medium. In a dielectric medium total charge consists of free charge. If p is the charge density of free charge at a. point in a small volume element dV. Then Gauss’s law can be expresses as
“The total normal electrical induction over a closed surface is equal to – times of 1/ ε 0total charge enclosed.
where p = charge per unit volume
V = volume enclosed by charge.
By Gauss transformation formula
ʃV div E dv =1/ε0 ʃ v p dv [ʃ s A n ds = ʃ v div Adv]
div E =1/ε0 P
ε0 div E =P
div ε0 E =P
P=0 Then D =ε0 E [D =ε0 E+P]
2. Derivation of Second Equation
div B = ∆.B = 0
“It is nothing but the differential form of Gauss law of magnetostatics.”
Since isolated magnetic poles and magnetic currents due to them have no significance.Therefore magnetic lines of force in general are either closed curves or go off to infinity.Consequently the number of magnetic lines of force entering any arbitrary closed surface isexactly the same as leaving it. It means that the flux of magnetic induction B across anyclosed surface is always zero.
Gauss law of magnetostatics states that “Total normal magnetic induction over aclosed surface is equal to zero.”
i. e; ʃ S B n ds =0
Applying Gauss transformation formula we get
ʃ V div B dv =0
The integrand should vanish for the surface boundary as the volume is arbitrary.
div B =0
3. Derivation of Third Law
Curl E =- Db /dt
“It is nothing but the differential form of Faraday’s law of electromagnetic induction.”
According to Faraday’s law of electromagnetic induction, it is known that e.m.f.induced in a closed loop is defined as negative rate of change of magnetic flux i.e.,
e=–dθ/dt
where θ = magnetic flux
or
ϕ = ʃ S BndS
ϕ=B / A where S is any surface having loop as boundary.
ʃ l E .dl = – d ϕ/dt
Putting the value of ϕ in equation (1), we get
ʃl E .dl = – d /dt ʃS b .n ds
ʃl E .dl = ʃS-dB /dt .n ds
Applying Stoke’s transformation formula on L.H.S.
ʃs( curl EndS= ʃS =Db /DT n ds
or ʃ S (curl E +dB /dt ) n ds =0
Further validity of the equation
Curl E =Db /dt =0
Or curl E = 0 Db /dt
This is known as Maxwell’s third equation.
4. Derivation of Maxwell’s Fourth Equation:
“This is nothing but differential form of modified Ampere circuital law.”
Ampere’s modified circuital law
According to law the work done in carrying a unit magnetic pole once around closed arbitrary path linked with the current is expressed by
ʃl B dl =µ 0 x i
i = current enclosed by the path
ʃl B dl =µ 0ʃs J n ds
On applying Stoke’s transformation formula in L.H.S.
ʃ s curl B n ds = ʃs µ 0 J n ds
è ʃ s (curl B- µ 0 J) n ds =0
For the validity this equation
curl B- µ 0 J =0
curl B- µ 0 J
It is known as the fourth equation of Maxwell.
Taking divergence of both sides
Div .(curl B )= div (µ 0 J)
0= div(µ 0 J)
0 div J [div (cual A) =0]
Div J =0
Which means that the current is always closed and there are no source and sink. Thuswe arrive at contradiction equation (3) is also in conflict with the equation of discontinuity.
But the according to law of continuity
Div J = – d p / d t
So this equation fails and it need of little modification. So Maxwell assume that
curl B = µ 0 (div J ) +µ 0(div J d)
0= µ0 (div J ) +µ 0(div J d)
By putting div J d =dp/dt
Div J d =div dD/ dt
Jd =Dd /dt(By Maxwell first equation, div D = p in equation (4))
Putting in equation (4), we get
Curl B =µ0(J +Dd /dt)
B =µ0 =H
Curl (µ0 =H) =µ0( J +Dd /dt)
Curl H =j dD /dt
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https://converter.ninja/volume/dry-gallons-to-imperial-cups/728-drygallon-to-brcup/
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# 728 dry gallons in imperial cups
## Conversion
728 dry gallons is equivalent to 11286.2011449349 imperial cups.[1]
## Conversion formula How to convert 728 dry gallons to imperial cups?
We know (by definition) that: $1\mathrm{drygallon}\approx 15.5030235507348\mathrm{brcup}$
We can set up a proportion to solve for the number of imperial cups.
$1 drygallon 728 drygallon ≈ 15.5030235507348 brcup x brcup$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brcup}\approx \frac{728\mathrm{drygallon}}{1\mathrm{drygallon}}*15.5030235507348\mathrm{brcup}\to x\mathrm{brcup}\approx 11286.201144934936\mathrm{brcup}$
Conclusion: $728 drygallon ≈ 11286.201144934936 brcup$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 imperial cup is equal to 8.86037726209393e-05 times 728 dry gallons.
It can also be expressed as: 728 dry gallons is equal to $\frac{1}{\mathrm{8.86037726209393e-05}}$ imperial cups.
## Approximation
An approximate numerical result would be: seven hundred and twenty-eight dry gallons is about eleven thousand, two hundred and eighty-six point two zero imperial cups, or alternatively, a imperial cup is about zero times seven hundred and twenty-eight dry gallons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
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Extension Principle — Concepts
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# Extension Principle — Concepts - PowerPoint PPT Presentation
Extension Principle — Concepts. To generalize crisp mathematical concepts to fuzzy sets . Extension Principle.
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Presentation Transcript
Extension Principle — Concepts
• To generalize crisp mathematical concepts to fuzzy sets.
Extension Principle
Extension Principle
• Let X be a cartesian product of universes X=X1…Xr, and be r fuzzy sets in X1,…,Xr, respectively. f is a mapping from X to a universe Y, y=f(x1,…,xr), Then the extension principle allows us to define a fuzzy set in Y by
where
Extension Principle
Example 1
f(x)=x2
Extension Principle
Fuzzy Numbers
• To qualify as a fuzzy number, a fuzzy set on R must possess at least the following three properties:
• must be a normal fuzzy set
• must be a closed interval for every α(0,1](convex)
• the support of , must be bounded
Extension Principle
Positive (negative) fuzzy number
• A fuzzy number is called positive (negative) if its membership function is such that
Extension Principle
Increasing (Decreasing) Operation
• A binary operation in R is called increasing (decreasing) if
for x1>y1 and x2>y2
x1x2>y1y2(x1x2<y1y2)
Extension Principle
Example 2
• f(x,y)=x+y is an increasing operation
• f(x,y)=x•y is an increasing operation on R+
• f(x,y)=-(x+y) is an decreasing operation
Extension Principle
Notation of fuzzy numbers’ algebraic operations
• If the normal algebraic operations +,-,*,/ are extended to operations on fuzzy numbers they shall be denoted by
Extension Principle
Theorem 1
• If and are fuzzy numbers whose membership functions are continuous and surjectivefromR to [0,1] and is a continuous increasing (decreasing) binary operation, then is a fuzzy number whose membership function is continuous and surjective from R to [0,1].
Extension Principle
Theorem 2
• If , F(R) (set of real fuzzy number) with and continuous membership functions, then by application of the extension principle for the binary operation : R R→R the membership function of the fuzzy number is given by
Extension Principle
Special Extended Operations
• If f:X→Y, X=X1 the extension principle reduces for all F(R) to
Extension Principle
Example 31
• For f(x)=-x the opposite of a fuzzy number is given with , where
• If f(x)=1/x, then the inverse of a fuzzy number is given with
, where
Extension Principle
Example 32
• For λR\{0} and f(x)=λx then the scalar multiplication o a fuzzy number is given by , where
Extension Principle
• Since addition is an increasing operation→ extended addition of fuzzy numbers that
is a fuzzy number — that is
Extension Principle
Properties of
• ( )( )
• is commutative
• is associative
• 0RF(R) is the neutral element for , that is , 0= , F(R)
• For there does not exist an inverse element, that is,
Extension Principle
Extended Product
• Since multiplication is an increasing operation on R+ and a decreasing operation on R-, the product of positive fuzzy numbers or of negative fuzzy numbers results in a positive fuzzy number.
• Let be a positive and a negative fuzzy number then is also negative and results in a negative fuzzy number.
Extension Principle
(
)
(
)
=
1=
1=
Properties of
• is commutative
• is associative
• , 1RF(R) is the neutral element for , that is , ,F(R)
• For there does not exist an inverse element, that is,
Extension Principle
Theorem 3
• If is either a positive or a negative fuzzy number, and and are both either positive or negative fuzzy numbers then
Extension Principle
Extended Subtraction
• Since subtraction is neither an increasing nor a decreasing operation,
• is written as ( )
Extension Principle
Extended Division
• Division is also neither an increasing nor a decreasing operation. If and are strictly positive fuzzy numbers then
The same is true if and are strictly negative.
Extension Principle
={(2,0.3),(3,0.3),(4,0.7),(6,1),(8,0.2),(9,0.4),(12,0.2)}
Note
• Extended operations on the basis of min-max can’t directly applied to “fuzzy numbers” with discrete supports.
• Example
• Let ={(1,0.3),(2,1),(3,0.4)}, ={(2,0.7),(3,1),(4,0.2)} then
No longer be convex → not fuzzy number
Extension Principle
Extended Operations for LR-Representation of Fuzzy Sets
• Extended operations with fuzzy numbers involve rather extensive computations as long as no restrictions are put on the type of membership functions allowed.
• LR-representation of fuzzy sets increases computational efficiency without limiting the generality beyond acceptable limits.
Extension Principle
Definition of L (and R) type
• Map R+→[0,1], decreasing, shape functions if
• L(0)=1
• L(x)<1, for x>0
• L(x)>0 for x<1
• L(1)=0 or [L(x)>0, x and L(+∞)=0]
Extension Principle
Definition of LR-type fuzzy number1
• A fuzzy number is of LR-type if there exist reference functions L(for left). R(for right), and scalars α>0, β>0 with
Extension Principle
Definition of LR-type fuzzy number2
• m; called the mean value of , is a real number
• α,β called the left and right spreads, respectively.
• is denoted by (m,α,β)LR
Extension Principle
Example 4
• Let L(x)=1/(1+x2), R(x)=1/(1+2|x|), α=2, β=3, m=5 then
Extension Principle
Fuzzy Interval
• A fuzzy interval is of LR-type if there exist shape functions L and R and four parameters , α, β and the membership function of is
The fuzzy interval is denoted by
Extension Principle
Different type of fuzzy interval
• is a real crisp number for mR→
=(m,m,0,0)LR L, R
• If is a crisp interval, →
=(a,b,0,0)LRL, R
• If is a “trapezoidal fuzzy number”→ L(x)=R(x)=max(0,1-x)
Extension Principle
Theorem 4
• Let , be two fuzzy numbers of LR-type: =(m,α,β)LR, =(n,γ,δ)LR Then
• (m,α, β)LR(n, γ,δ)LR=(m+n, α+γ, β+δ)LR
• -(m, α, β)LR=(-m, β, α)LR
• (m, α, β)LR (n, γ, δ)LR=(m-n, α+δ, β+γ)LR
Extension Principle
Example 5
• L(x)=R(x)=1/(1+x2)
• =(1,0.5,0.8)LR
• =(2,0.6,0.2)LR
• =(3,1.1,1)LR
• =(-1,0.7,1.4)LR
Extension Principle
Theorem 5
• Let , be fuzzy numbers →
(m, α, β)LR (n, γ, δ)LR≈(mn,mγ+nα,mδ+nβ)LR for , positive
• (m, α, β)LR (n, γ, δ)LR≈(mn,nα-mδ,nβ-mγ)LR for positive, negative
• (m, α, β)LR (n, γ, δ)LR ≈(mn,-nβ-mδ,-nα-mγ)LR for , negative
Extension Principle
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https://testbook.com/question-answer/an-object-with-a-specific-mass-will-weigh-______--63aeca76b382478bee323c4a
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# An object with a specific mass will weigh ______.
This question was previously asked in
RRB ALP Electrician 22 Jan 2019 Official Paper (Shift 3)
View all RRB ALP Papers >
1. zero on the Earth
2. less on the Earth than on the Moon
3. more on the Earth than on the Moon
4. equal on the Earth and the moon
## Answer (Detailed Solution Below)
Option 3 : more on the Earth than on the Moon
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## Detailed Solution
Concept:
Weight:
• It is the measure of the force of gravity acting on a body.
• Formula: Weight, W = mg, where m = mass, a = acceleration
• So it can be said that the weight of an object is directly proportional to its mass.
• As weight is a force its SI unit is also the same as that of force, SI unit of weight is Newton (N).
Acceleration due to gravity:
• It is the acceleration gained by an object due to gravitational force.
• Its SI unit is m/s2.
• It has both magnitude and direction, hence, it’s a vector quantity.
• Acceleration due to gravity is represented by g.
• The standard value of g on the surface of the earth at sea level is 9.8 m/s2.
• The acceleration on the moon is one-sixth of the acceleration due to gravity on the earth.
Explanation:
Weight, W = mg
On the moon, g' = g/6, where, g = acceleration due to gravity on the surface of earth.
• Hence, the weight is directly proportional to the acceleration due to gravity.
• The acceleration due to gravity on the moon is less as compared to the surface of earth.
• Therefore, the weight on the surface of earth is more than that on the moon's surface.
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https://allclearmister.com/what-is-cv-in-flow-rate/
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What is CV in flow rate?
What is CV in flow rate?
Valve Flow Coefficient (Cv) is a valve’s capacity for a liquid or gas to flow through it. It is technically defined as the volume of water at 60F (in US gallons) that will flow through a valve per minute with a pressure drop of 1 psi across the valve.
What is CV rating?
The Cv rating for each valve is listed in the tables found in the valve catalog. The definition of Cv is the # of gallons of water that will flow through the valve with a 1 PSI pressure differential when the valve is open. The equations below can be used to determine: Flow Rate, given the Cv and P.
How do you calculate pressure drop in a CV?
Calculation of Pressure Loss using the Cv MethodC v = Q S G P \\displaystyle C_v = Q \\sqrt{\\frac{SG}{\\Delta P}} Cv=QPSG.C v = 0.0694 Q P 999 \\displaystyle C_v = 0.0694 Q \\sqrt{ \\frac{\\rho}{\\Delta P \\times 999}} Cv=0. 0694QP999K v = Q S G P \\displaystyle K_v = Q \\sqrt{ \\frac{SG}{ \\Delta P}} Kv=QPSG.
How many GPM can a 1 pipe flow?
Assume Average Pressure. (20-100PSI) About 12f/s flow velocity1″1.00-1.03″37 gpm1.25″1.25-1.36″62 gpm1.5″1.50-1.60″81 gpm2″1.95-2.05″127 gpm9
How do you calculate flow rate per hour?
If you simply need to figure out the mL per hour to infuse, take the total volume in mL, divided by the total time in hours, to equal the mL per hour. For example, if you have 1000 mL NS to infuse over 8 hours, take 1000 divided by 8, to equal 125 mL/hr. To calculate the drops per minute, the drop factor is needed.
How is IVPB rate calculated?
1:53Suggested clip 113 secondsIVPB infusion – YouTubeYouTubeStart of suggested clipEnd of suggested clip
How is flow rate calculated in pharmacy?
3:46Suggested clip 108 secondsHow to solve flow rate calculations problems – YouTubeYouTubeStart of suggested clipEnd of suggested clip
How many drops is 100 ml per hour?
Reference Chart of Drops per MinuteIV Tubing Drop FactorDesired Hourly Rate: ML / HR2010010 DROP/ML31615 DROP/ML52520 DROP/ML6322
How do you calculate IV fluid drop rate?
Drip Rates — is when the infusion volume is calculated into drops. The formula for the Drip Rate: Drip Rate = Volume (mL) Time (h) . A patient is ordered to receive 1 000 mL of intravenous fluids to run over 8 hours.
How many drops GTTS are in 1 mL?
20 drops
How do you calculate GTT mL?
Example of calculating gtts/min:Order: 1000 mLof D5/W to infuse 130 mL /hr.Drop factor of tubing is 20 gtts = 1 mL.Formula:mL/hr X drop factor = gtt/min.60 minutes.130mL/hr x 20 gtt/min = 2600 = 43 gtt/min.
What is GTT mL?
In hospitals, intravenous tubing is used to deliver medication in drops of various sizes ranging from 10 drops/mL to 60 drops/mL. A drop is abbreviated gtt, with gtts used for the plural, often seen on prescriptions. These abbreviations come from gutta (plural guttae), the Latin word for drop.
What is drop rate?
The drop rate is the frequency at which a monster is expected to yield a certain item when killed by players. When calculating a drop rate, divide the number of times you have gotten the certain item, by the total number of that monster that you have killed. Bones have a 100% drop rate from chickens.
How do you calculate drugs?
6:17Suggested clip 115 secondsDrug Calculations – basic examples 105 – YouTubeYouTubeStart of suggested clipEnd of suggested clip
How do nurses calculate drugs?
Determine in which units your drug is measured (units/hour, mg/hour, or mcg/kg/minute). Know the patient’s weight in kg if your calculation is weight based. Use the universal formula below and then divide your final answer by the patient’s weight in kg to arrive at mcg/kg/minute.
What is a drug calculation test?
A drug calculation test is an examination of your ability to efficiently work through the mental arithmetic associated with administering drugs to patients. You are working with dosage formulas, weights and volume, translating different types of measurements and working out dosage schedules.
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A025129 a(n) = p(1)p(n) + p(2)p(n-1) + ... + p(k)p(n-k+1), where k = [ n/2 ], p = A000040, the primes. 16
0, 6, 10, 29, 43, 94, 128, 231, 279, 484, 584, 903, 1051, 1552, 1796, 2489, 2823, 3784, 4172, 5515, 6091, 7758, 8404, 10575, 11395, 14076, 15174, 18339, 19667, 23414, 24906, 29437, 31089, 36500, 38614, 44731, 47071, 54198, 56914, 65051, 68371, 77402, 81052, 91341 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS This is the sum of distinct squarefree semiprimes with prime indices summing to n + 1. A squarefree semiprime is a product of any two distinct prime numbers. A prime index of n is a number m such that the m-th prime number divides n. The multiset of prime indices of n is row n of A112798. - Gus Wiseman, Dec 05 2020 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 Gus Wiseman, Sum of prime(i) * prime(j) for i + j = n, i != j. FORMULA a(n) = A024697(n) for even n. - M. F. Hasler, Apr 06 2014 EXAMPLE From Gus Wiseman, Dec 05 2020: (Start) The sequence of sums begins (n > 1): 6 = 6 10 = 10 29 = 14 + 15 43 = 22 + 21 94 = 26 + 33 + 35 128 = 34 + 39 + 55 231 = 38 + 51 + 65 + 77 279 = 46 + 57 + 85 + 91 (End) MATHEMATICA f[n_] := Block[{primeList = Prime@ Range@ n}, Total[ Take[ primeList, Floor[n/2]]*Reverse@ Take[ primeList, {Floor[(n + 3)/2], n}]]]; Array[f, 44] (* Robert G. Wilson v, Apr 07 2014 *) PROG (PARI) A025129=n->sum(k=1, n\2, prime(k)*prime(n-k+1)) \\ M. F. Hasler, Apr 06 2014 (Haskell) a025129 n = a025129_list !! (n-1) a025129_list= f (tail a000040_list) [head a000040_list] 1 where f (p:ps) qs k = sum (take (div k 2) \$ zipWith (*) qs \$ reverse qs) : f ps (p : qs) (k + 1) -- Reinhard Zumkeller, Apr 07 2014 CROSSREFS Cf. A000040, A258323. The nonsquarefree version is A024697 (shifted right). Row sums of A338905 (shifted right). A332765 is the greatest among these squarefree semiprimes. A001358 lists semiprimes. A006881 lists squarefree semiprimes. A014342 is the self-convolution of the primes. A056239 is the sum of prime indices of n. A338899/A270650/A270652 give the prime indices of squarefree semiprimes. A339194 sums squarefree semiprimes grouped by greater prime factor. Cf. A001221, A005117, A062198, A098350, A168472, A320656, A338900, A338901, A338904, A339114, A339116. Sequence in context: A240972 A349846 A103767 * A093559 A269697 A271067 Adjacent sequences: A025126 A025127 A025128 * A025130 A025131 A025132 KEYWORD nonn AUTHOR EXTENSIONS Following suggestions by Robert Israel and N. J. A. Sloane, initial 0=a(1) added by M. F. Hasler, Apr 06 2014 STATUS approved
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# Statistics
posted by .
In a certain city, there are 10,000 persons age 18 to 24. A simple random sample of
500 such persons is drawn, of whom 200 turn out to be currently enrolled in college. Find a 90%
con¯dence interval for the percentage of persons age 18 to 24 in the city who are attending college.
• Statistics -
Try using a binomial proportion CI formula:
CI90 = p + or - 1.645(√pq/n)
...where p = proportion in the problem, q = 1 - p, n = sample size, √ = square root, and + or - 1.645 represents 90% CI using a z-table.
p = 200/500
q = 1 - p
n = 500
Convert all fractions to decimals. Substitute values into the formula and determine the confidence interval.
I hope this will help get you started.
• Statistics -
500.12
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'
# Search results
Found 1194 matches
Roller screw (effective nut inside diameter)
A roller screw, also known as a planetary roller screw or satellite roller screw, is a low-friction precision screw-type actuator, a mechanical device for ... more
Roller screw ( effective roller diameter)
A roller screw, also known as a planetary roller screw or satellite roller screw, is a low-friction precision screw-type actuator, a mechanical device for ... more
Worksheet 306
Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.
(a) The figure shows the forearm of a person holding a book. The biceps exert a force FB to support the weight of the forearm and the book. The triceps are assumed to be relaxed. (b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint
Strategy
There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .
Solution
The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium (net τ = 0) becomes
Force (Newton's second law)
Torque
Force (Newton's second law)
Torque
Note that sin θ = 1 for all forces, since θ = 90º for all forces. This equation can easily be solved for FB in terms of known quantities,yielding. Entering the known values gives
Mechanical equilibrium - 3=3 Torque example
which yields
Torque
Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is
Division
Discussion
This means that the biceps muscle is exerting a force 7.38 times the weight supported.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Speed of sound in three-dimensional solids (pressure waves)
The speed of sound is the distance travelled per unit of time by a sound wave propagating through an elastic medium. Sound travels faster in liquids and ... more
Creep (deformation)
In materials science, creep (sometimes called cold flow) is the tendency of a solid material to move slowly or deform permanently under the influence of ... more
Young - Laplace equation
In physics, the Young – Laplace equation, is a nonlinear partial differential equation that describes the capillary pressure difference sustained ... more
Turnbuckle (The direct shear stress induced in screw thread)
A mechanical joint is a part of a machine which is used to connect another mechanical part or mechanism. Mechanical joints may be temporary or permanent. ... more
Friction Loss (hydraulic slope) - related to pressure change
In fluid flow, friction loss (or skin friction) is the loss of pressure or “head” that occurs in pipe or duct flow due to the effect of the fluid’s ... more
Lift-to-Drag Ratio
In aerodynamics, the lift-to-drag ratio, or L/D ratio, is the amount of lift generated by a wing or vehicle, divided by the drag it creates by moving ... more
Force aplied at a contact area between a sphere and an elastic half-space
Contact mechanics is the study of the deformation of solids that touch each other at one or more points. Hertzian contact stress refers to the localized ... more
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0
# How many 70 kg of coffee bags can you fit into 40 foot long container?
Updated: 9/26/2023
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Q: How many 70 kg of coffee bags can you fit into 40 foot long container?
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[object Object]
### What is a good way to store coffee for a long period of time?
Good ways to store coffee long term is to keep the coffee in an air tight container away from heat. You can also store coffee in vacuum sealed bags in the freezer.
### How many 50 pound bags can you fit into a 40 foot long container?
That depends on the dimensions of the bags. You can't equate weight to area or volume. A 50 pound bag of feathers is a lot bigger than a 50 pound bag of sand.
### Can I mail coffee?
yeah as long it is in a container that is 100% seal-able in a carton
### What does ship TEU mean?
Twenty-foot Equivalent Unit. A container that is 20 feet long is 1 TEU while a 40 foot container unit is 2 TEU. A 53 foot container is 2.65 TEUs
40
20
### How many barrels can fit in a 20 foot long x 8 foot wide container?
How big are the barrels? You are missing the third part of the equation.
### How many square foot is in 40 foot container?
The square feet are simply the width times the length. But, I will assume you are interested in the inside dimension of the container or the volume. A sea container is 39' and 3/8" long by 7' 8-3/8"wide by 7' 9-5/8" high. Therefore a 40 foot container holds 2261 cubic feet of area
yes
### A container 10 foot long 5 foot wide 2 foot deep how many liters?
The volume 10 foot times 5 foot times 2 foot is 100 cubic feet. 100 cubic feet are 2831.6846592 liters.
### How do you store coffee?
Coffee can last a long time when stored properly, and the best place to store it would be in a cool, dark place. Make sure that the top is sealed well too so that air doesn't get in to the container.
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Hypothesis Testing (Chi-Squared Test) (G Dataflow)
Tests hypotheses about the variance of a population whose distribution is at least approximately normal.
sample set
Randomly sampled data from the population.
variance
Hypothesized variance of the population.
The null hypothesis is that the population variance is equal to variance.
Default: 1
significance level
Probability that this node incorrectly rejects a true null hypothesis.
Default: 0.05
error in
Error conditions that occur before this node runs.
The node responds to this input according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Default: No error
alternative hypothesis
Hypothesis to accept if this node rejects the null hypothesis that the population variance is equal to variance.
Name Value Description
variance(pop) != variance 0 The population variance is not equal to variance.
variance(pop) > variance 1 The population variance is greater than variance.
variance(pop) < variance -1 The population variance is less than variance.
Default: variance(pop) != variance
null hypothesis rejected?
A Boolean that indicates whether this node rejects the null hypothesis.
True p value is less than or equal to significance level. This node rejects the null hypothesis and accepts the alternative hypothesis. False p value is greater than significance level. This node accepts the null hypothesis and rejects the alternative hypothesis.
p value
Smallest significance level that leads to rejection of the null hypothesis based on the sample sets.
confidence interval
Lower and upper limits for the population variance. confidence interval indicates the uncertainty in the estimate of the true population variance.
low
Lower limit of the estimate of the population variance.
high
Upper limit of the estimate of the population variance.
chi-squared test information
Sample statistics of the chi-squared test.
sample variance
Variance of sample set.
degree of freedom
Degree of freedom of the chi-squared distribution that the test statistic follows.
sample chi-squared value
Sample test statistic used in the chi-squared test.
sample chi-squared value is equal to $\frac{\left(n-1\right)\text{\hspace{0.17em}}*\text{\hspace{0.17em}}\mathrm{sample variance}}{\mathrm{variance}}$.
chi-squared critical value (lower)
Lower chi-squared value that corresponds to significance level and alternative hypothesis.
Algorithm for Calculating chi-squared critical value (lower)
Let Xn represent a chi-squared distributed variate with n degrees of freedom. chi-squared critical value (lower) satisfies the following equations based on the value of alternative hypothesis.
alternative hypothesis chi-squared critical value (lower)
variance(pop) != variance Prob{Xn < chi-squared critical value (lower)} = significance level / 2
variance(pop) > variance chi-squared critical value (lower) = NaN
variance(pop) < variance Prob{Xn < chi-squared critical value (lower)} = significance level
chi-squared critical value (upper)
Upper chi-squared value that corresponds to significance level and alternative hypothesis.
Algorithm for Calculating chi-squared critical value (upper)
Let Xn represent a chi-squared distributed variate with n degrees of freedom. chi-squared critical value (upper) satisfies the following equations based on the value of alternative hypothesis.
alternative hypothesis chi-squared critical value (upper)
variance(pop) != variance Prob{Xn > chi-squared critical value (upper)} = significance level / 2
variance(pop) > variance Prob{Xn > chi-squared critical value (upper)} = significance level.
variance(pop) < variance chi-squared critical value (upper) = NaN
error out
Error information.
The node produces this output according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported
Web Server: Not supported in VIs that run in a web application
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# Difference between revisions of "1984 AIME Problems/Problem 14"
## Problem
What is the largest even integer that cannot be written as the sum of two odd composite numbers?
## Solution 1
Take an even positive integer $x$. $x$ is either $0 \bmod{6}$, $2 \bmod{6}$, or $4 \bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:
If $x \ge 18$ and is $0 \bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.
If $x\ge 44$ and is $2 \bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.
If $x\ge 34$ and is $4 \bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.
Clearly, if $x \ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.
## Solution 2
Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $\boxed{038}$.
## Solution 3 (bash)
Let $2n$ be an even integer. Using the Chicken McNugget Theorem on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as $n+n$. We bash each case until we find one that works.
## Solution 4 (easiest)
The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...
For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14.
If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\boxed{38}$
## Solution 5
Claim: The answer is $\boxed{038}$.
Proof: It is fairly easy to show 38 can't be split into 2 odd composites.
Assume there exists an even integer $m > 38$ that m can't be split into 2 odd composites.
Then, we can consider m modulo 5.
If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since $m > 20$, $k > 1$ so $k \geq 2$. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well.
If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since $m > 26$, $k > 1$ so $k \geq 2$. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well.
If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since $m > 32$, $k > 1$ so $k \geq 2$. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well.
If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since $m > 38$, $k > 1$ so $k \geq 2$. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well.
If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since $m > 14$, $k > 1$ so $k \geq 2$. Thus, 5k is composite. Since 9 is odd and composite, m is even, 5k should be odd as well.
Thus, in all cases we can split m into 2 odd composites, and we get at a contradiction. $\blacksquare$
-Alexlikemath
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Next: An example: Love Affairs Up: Lab 2: Life on Previous: Lab 2: Life on
## Introduction to 2-D systems: Basic concepts
In a 2-D system we will consider dynamical systems that look something like
where x and y are our two variables of interests.. Examples might include, rabbits and grass, hosts and parasites or maybe Romeo and Juliet (...see Below). The most important concepts to understand about 2-D systems (and general dynamical systems are)
• the phase plane
• Flow on the phase plane
• Phase portraits
• Fixed points
• Stability
The Phase plane is a graph where the axes are just our variables x and y, so instead of plotting rabbits against time and grass against time, we want to look at the behavior of rabbits against grass. If we had three variables, the volume that they span is known as phase space (beyond three variables, life gets tricky). Flow on the phase plane is exactly the same idea we used in constructing direction sets (Figure 1), that is, Equations (6) say that for every point x,y on the phase plane, and give rules for how the point will change in time. If is positive x will increase, if negative x will decrease. The same is true for y, so at every point there will be a little arrow saying where the system will go in a short time step. This is easier to show than say. Once we know what the flow field looks like, individual solutions simply trace out trajectories in phase space.
Now in general, where the change functions are not zero, the system will evolve in time along various trajectories. Things get more interesting however, around the few fixed points where things don't change. At a fixed point both and are zero so if we start exactly at a fixed point we will never move away from it. The more interesting question is what happens if we start close to a fixed point. Like in the 1-D problem we can have stable attractors and unstable repellers but in two dimensions we can do some other strange things as well. The next problem will illustrate what can happen. Nevertheless, the basic rule for analysis of a 2-D system is
1. formulate an interesting 2-D problem
2. find the fixed points and categorize their stability
3. sketch out a phase portrait
4. Use stella to solve for a few crucial trajectories
When you are done, you will have a lovely picture that tells you exactly how the entire system will evolve in time. Many times you can guess what will happen without even solving the equations. The next example shows how.
Next: An example: Love Affairs Up: Lab 2: Life on Previous: Lab 2: Life on
marc spiegelman
Mon Sep 22 21:30:22 EDT 1997
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# Stats Exam 2 *Focus*
The flashcards below were created by user Radhika316 on FreezingBlue Flashcards.
-when to use Mean?
• -Takes all the observation/scores into account; highest and lowest scores are accounted for
• -Takes the distance & direction of deviations/errors into account.
2. More uses than median & mode: Necessary for calculating many inferential statistics.
3. ONLY Use the mean for unimodal, symmetrical distributions of interval/ratio level data.
2. Limitations of Mean:
• 1. Not always possible to calculate a mean (scale); Mean can only be calculated for interval/ratio level data
• -Need a different measure for nominal or ordinal level data
• 2. Not always appropriate to use the mean to describe the middle of a distribution (distribution)
• -Mean is sensitive to extreme values or “outliers”
• -Mean does not always reflect where the scores “pile up”
• -Need a different measure for asymmetrical distributions: positive or negative distribution
3. **When to use the Mean:
Use the mean for unimodal, symmetrical distributions of interval/ratio level data.
4. THE MEDIAN:
• -The middle score.
• -the goal is to find the midpoint "score" of the distribution; it can either equal a score or be the point between two middle scores
• -Calculate by: Order the scores from least to greatest, find middle score or avg the two middle scores
• -Divides the distribution exactly in half; 50th percentile
• -Odd # of scores & no “pileup” or ties at the middle.
-which scales can it be used with?
-what position does it have in skewed distributions
1. Insensitive to extreme values; Can be used when extreme values distort the mean
2. The most central, representative value in skewed distributions
3. Can be calculated when the mean cannot
4.Can be used with ranks/ordinal scales (as well as interval/ratio data)
5. Can be used with open-ended distributions; Example: # of siblings (5+ siblings?)
6. Limitation of The Median:
-which scale can't be used?
-informative ?
• 1. Not as informative as the mean; just cuts the scores in half
• 2. Takes only the observations/scores around the 50th percentile into account.
• 3.Provides no information about distances between observations.
• 4.Fewer uses than the mean; Median is purely descriptive (rather than informative)
• 5.Not always possible to calculate a median (scale); can only be calculated for ordinal & interval/ratio data
7. When to use the median:
Use the median when you cannot calculate a mean or when the distributions of interval/ratio data are skewed by extreme values; ordinal scales, open ended distributions
8. THE MODE:
The most frequently occurring score(s); describes where the scores pile up.
-which scales? how does this compare with mean/median?
1.Simple to find
2.Can be used with ANY scale of measurement; Median can only be calculated for ordinal & interval/ratio data vs. Mean can only be calculated for interval/ratio data
3. Can be used to indicate >1 most frequent value; Use to indicate bimodality, multimodality
10. Limitations of the Mode:
• 1. Not as informative as the mean or median;
• -Takes only the most frequently observed X values into account.
• -Provides no information about distances between observations or the # of observations above/below the mode.
• 2. Fewer uses than the mean: Mode is purely descriptive.
• -Need to calculate a mean to use with inferential statistics.
11. When to use Mode:
• -Use the mode when you cannot compute a mean or median,
• -or with the mean/median to describe a bimodal/multimodal distribution
• Note:
• Symmetrical: Mean=median=mode
• Bimodial: mean & median is in the middle
• -nominal scale,
• -to get shape of the distribution.
12. Variability measures used with the mode & median--> Range:
-Used for what categories?
-what scales?
-when is range used?
Range (For Mode): Based on the distance between the highest & lowest observations on the X scale. Only takes the 2 most extreme observations into account.
For interval/ratio data, range = highest score - lowest score
--Range can also be used for ordered categories: Ex:: from “agree” to “disagree strongly,” with modal response = “disagree.”
--The range is typically used with the mode, when the mean & median are inappropriate or impossible to calculate (but may be reported along with a median or a mean).
13. Variability measures used with the mean
-used with what scales?
-most useful for what kind of distribution?
-SS, variance & standard deviation are based on distances between each of the scores & the mean on the X scale. All scores are taken into account, as with the mean.
-Use only with interval/ratio data.
-Most useful for symmetrical distributions, when the mean is the best measure of central tendency.
Author: Radhika316 ID: 266276 Card Set: Stats Exam 2 *Focus* Updated: 2014-03-13 10:52:18 Tags: Stats Folders: 2014 Description: Mode, Mean, Median, Range... Show Answers:
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# 2013 AMC 12B Problems/Problem 16
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$
## Solution 1
The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure $180^\circ - 108^\circ = 72^\circ$. The base angles are equal, so the triangles must be isosceles.
Let one of the sides of the pentagon have length $x_1$ (and the others $x_2, x_3, x_4, x_5$). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length $\frac{x_1}{2} \sec 72^\circ$, and so the two sides together have length $x_1 \sec 72^\circ$. To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get $(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ$ (because the perimeter of the pentagon is $1$). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is $\boxed{\textbf{(A)} \ 0}$.
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# Under which angle conditions could a triangle exist? Check all that apply.
Step-by-step explanation: We have to check under which condition a triangle will exist. According to angle sum property of the triangle, the sum of all the three angles of a triangle is 180 degrees.An acute angle is an angle with a a measure less than 90 degreesA right angle is an angle with a measure of 90 degrees.An obtuse angle is an angle with a measure of greater than 90 degrees. 1. It is possible to have triangle with three acute angle Example: A triangle with all the three angles of 60 degrees 2. It is possible to have a triangle with 2 acute angles and 1 right angle. Example: A triangle with all the two angles of 45 degrees and one right angle. 3. It is not possible to have a triangle with 1 acute angle, 1 right angle and 1 obtuse angle. It will violate the angle sum property of triangle. 4. It is not possible to have a triangle with 1 acute angle and 2 obtuse angles. It will violate the angle sum property of triangle. 5. It is possible to have a triangle with 2 acute angles and 1 obtuse angle. Example: A triangle with all the angles of 45 degrees, 40 degrees and 95 degrees
3 acute angles and 2 acute angles, 1 obtuse angle and 2 acute angles, 1 right angle Step-by-step explanation: the key is just to imagine it also you have to know that right angle is 90° acute is less than 90° and obtuse is more than 90°
therefore three acute angles, eg. 60, 60 and 60 degrees can make up a triangle 2 acute angles and a right angle, say 45 and 45 degrees and 90 degrees make a triangle any obtuse angle + a right angle will be larger than 180 degrees, so no triangle can exist. another acute angle will make the number of degrees even larger any 2 obtuse angles will make a number larger than 180 degrees, so with an acute angle is impossible 2 acute and an obtuse angle is possible, say 40, 40 and 100 degrees
3 acute angles
2 acute angles, 1 right angle
2 acute angles, 1 obtuse angle Or 1, 2 , 5 Step-by-step explanation:
abe Step-by-step explanation:
Triangles DON’T exist for:1 acute angle, 1 right angle, 1 obtuse angle1 acute angle, 2 obtuse angles
3 acute angles
2 acute angles, 1 right angle
2 acute angles, 1 obtuse angle
Further Explanation The basic properties of triangles include
In a triangle, all the sum of the angle is 180 degree and it is also referred to as angle sum property.
In a triangle, the sum of the two sides is greater than the sum of the third side.
In a triangle, the longest side is the side that opposite the largest angle and vice versa. A triangle refers to a closed figure with a three-line segment and three angles. The three types of a triangle based on size are:
Equilateral triangle
, Isosceles triangle and Scalene triangle
An equilateral triangle is a triangle where the lengths of all the three sides are equal
An isosceles triangle is a triangle where its two sides are equal
A scalene triangle is a triangle that has three sides with different length However, the three types of a triangle based on angles include:
Acute-angled triangle: in Acute-angled triangle, all its angle are acute
Obtuse-angled triangle: in Obtuse-angled triangle, one its angle is obtuse
Right-angled triangle: in the Right-angled triangle, one of its angles is a right angle. Therefore, the correct answer is 3 acute angles, 2 acute angles, 1 right angle, 2 acute angles, 1 obtuse angle
LEARN MORE: Under which angle conditions could a triangle exist? Check all that apply. 1/2 of an obtuse angle is a(n): A. Obtuse angle. B. Acute angle KEYWORDS: right-angled triangleacute angles2 acute anglesobtuse anglea scalene triangle
3 acute angles. 2 acute angles, 1 right angle. 2 acute angles, 1 obtuse angle. Step-by-step explanation: The 3 angles in a triangles add up to 180 degrees. Acute angles are < 90 degrees. A right angle = 90 degrees, Obtuse angles are > 90 degrees.
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# Over the past twenty-five years the introduction of
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10 Mar 2010, 11:18
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Over the past twenty-five years the introduction of labor-saving technologies has greatly reduced the average amount of time a worker needs to produce a given output, potentially both reducing the number of hours each worker works each week and increasing workers’ leisure time correspondingly. The average amount of leisure time per worker, however, has increased at only half the rate at which the average hourly output per worker has grown.
If the statements above are true, which one of the following is most strongly supported by them?
(A) Workers on average spend more money on leisure activities today than they did twenty-five years ago.
(B) Labor-saving technologies have created fewer jobs than they have eliminated.
(C) The percentage of the population that is in the work force has grown over the past twenty-five years.
(D) The average hourly output per worker has not risen as much as had been anticipated when modern labor-saving technologies were first introduced.
(E) Twenty-five years ago the average weekly output per worker was less than it is today.
[Reveal] Spoiler: OA
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10 Mar 2010, 11:35
E
A B & C are out of Scope
D is not correct (as the stimuli doesn't mention Anticipation any where)
Left with E.
Even if you don't use elimination, only E seems to be in line with the stimuli, all others seem out of context.
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10 Mar 2010, 12:19
1
KUDOS
IMO it is E.
A) Workers on average spend more money on leisure activities today than they did twenty-five years ago.
>> Money out of scope.
(B) Labor-saving technologies have created fewer jobs than they have eliminated.
>> Less jobs out of scope.
(C) The percentage of the population that is in the work force has grown over the past twenty-five years.
>> Number of people is out of scope.
(D) The average hourly output per worker has not risen as much as had been anticipated when modern labor-saving technologies were first introduced.
>> At first I thought this could be one. But the passage mentions "twenty-five years the introduction of labor-saving technologies has greatly reduced the average amount of time a worker needs to produce a given output" .. that means it still is working. So D can not be the answer.
(E) Twenty-five years ago the average weekly output per worker was less than it is today.
>> Yes it is. This will be stongly supported by the above argument.
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10 Mar 2010, 21:12
OA is indeed E.
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07 Jul 2010, 01:57
E.
Guys , but what is interesting here so much ? )
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18 Feb 2011, 08:54
The stimulus,however,talks about numbers and averages and instinctively you think in terms of number and percentages chapter in Powerscore, the answer choices seems pretty easy and ONLY E is left out as every other AC seems out of scope.
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18 Feb 2011, 14:47
+1 E
Remember that Must be true questions don't accept new information in the answer choices.
The correct answer must be an inference based on the estimulus.
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21 Feb 2011, 00:57
shd be E
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21 Feb 2011, 02:54
Easy E.
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21 Feb 2011, 04:28
+1 for E
no other options are even remotely related.
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angel2009 wrote:
Over the past twenty-five years the introduction of labor-saving technologies has greatly reduced the average amount of time a worker needs to produce a given output, potentially both reducing the number of hours each worker works each week and increasing workers’ leisure time correspondingly. The average amount of leisure time per worker, however, has increased at only half the rate at which the average hourly output per worker has grown.
If the statements above are true, which one of the following is most strongly supported by them?
(A) Workers on average spend more money on leisure activities today than they did twenty-five years ago.
(B) Labor-saving technologies have created fewer jobs than they have eliminated.
(C) The percentage of the population that is in the work force has grown over the past twenty-five years.
(D) The average hourly output per worker has not risen as much as had been anticipated when modern labor-saving technologies were first introduced.
(E) Twenty-five years ago the average weekly output per worker was less than it is today.
only E makes sense.
A - out of scope
B - not discussed so no
C - not discussed so no
D - what was anticipated is out of scope.
E - last sentence says that the average hourly output per worker has grown.
E it is.
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Over the past twenty-five years the introduction of [#permalink]
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10 Jun 2016, 04:37
25 YEARS AGO :-Total hours available to a worker each week = 7 Days X 24 Hours = 168 Hours
Off course the worker will not work for all 24 hours a day. He would finish his job and relax. Those Relaxing hours are called Leisure hours.
Mathematically speaking :- Total hours in a week (Work hours in a week + Leisure hours in a week)
Lets assume 25 YEARS AGO worker works all 7 days for 10 hours a day (70 hours per week) and his rate was 10 toys an hour==> 10 toys per hour X 70 hours a week = 700 toys per week
Total weekly hours available= 168 hours
His weekly work hours = 70 hours
Weekly leisure time available to worker = (168-70)= 98 Hours
25 years ago Weekly output ===> 700 toys per week.
Now statements tells us that TODAY weekly output has grown by X amount But leisure time has grown by only X/2 amount.
Since the weekly production has definitely increased, lets assume it has increased 10 times. (it can be 2 times, 200 times, 1.8 times.. How many times it doesn't matter but the sure thing is that "hourly" output has increased and by extension the "weekly" output has also increased) so now the worker makes 10 times of 700 toys = 10 X 700==> 7000 toys
But his leisure time has increased only half the original leisure time= 98 hours + half of 98 hours= 98+49=147 hours (This statement is not totally useless by the way, as it will be used to calculate TODAYS working hours )
Today Worker works 168-147=21 hours .. WHATTTTT !! Today the worker works only 21 hours in a whole week and still produces 7000 toys a week
His weekly Production ===> 7000 TOYS per week
So , we can see that 25 years ago a worker produced 700 toys and today he produces 7000 toys. It means 25 years ago his weekly production was less compared to today.
What answer choice is reconfirming the math. It's E
(E) Twenty-five years ago the average weekly output per worker was less than it is today.
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Re: Over the past twenty-five years the introduction of [#permalink]
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07 Nov 2016, 08:40
(E) Twenty-five years ago the average weekly output per worker was less than it is today.
it might be that workers are working for longer hours twenty five years ago, so may be output worker may be equal to or larger 25 years ago.
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Re: Over the past twenty-five years the introduction of [#permalink] 07 Nov 2016, 08:40
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Home | | Cryptography and Network Security | Polynomial Arithmetic
Polynomial Arithmetic
Polynomial arithmetic in which the coefficients are in GF(p), and the polynomials are defined modulo a polynomial m(x) whose highest power is some integer n.
POLYNOMIAL ARITHMETIC
Before continuing our discussion of finite fields, we need to introduce the interest- ing subject of polynomial arithmetic. We are concerned with polynomials in a single variable x, and we can distinguish three classes of polynomial arithmetic.
Ordinary polynomial arithmetic, using the basic rules of algebra.
Polynomial arithmetic in which the arithmetic on the coefficients is performed modulo p ; that is, the coefficients are in GF(p).
Polynomial arithmetic in which the coefficients are in GF(p), and the polynomials are defined modulo a polynomial m(x) whose highest power is some integer n.
This section examines the first two classes, and the next section covers the last class.
Ordinary Polynomial Arithmetic
A polynomial of degree n (integer n >= 0) is an expression of the form
where the ai are elements of some designated set of numbers S, called the coefficient set, and an != 0. We say that such polynomials are defined over the coefficient set S.
A zero-degree polynomial is called a constant polynomial and is simply an element of the set of coefficients. An nth-degree polynomial is said to be a monic polynomial if an = 1.
In the context of abstract algebra, we are usually not interested in evaluating a
polynomial for a particular value of x [e.g., f(7)]. To emphasize this point, the variable
x is sometimes referred to as the indeterminate.
Polynomial arithmetic includes the operations of addition, subtraction, and multiplication. These operations are defined in a natural way as though the variable x was an element of S. Division is similarly defined, but requires that S be a field. Examples of fields include the real numbers, rational numbers, and Zp for p prime. Note that the set of all integers is not a field and does not support polynomial division.
Addition and subtraction are performed by adding or subtracting corresponding coefficients. Thus, if
In the last formula, we treat ai as zero for i > n and bi as zero for m. Note that the degree of the product is equal to the sum of the degrees of the two polynomials.
As an example, let f(x) = x3 + x2 + 2 and g(x) = x2 - x + 1, where S is the set of integers. Then
f(x) + g(x= x3 + 2x2 - x + 3
f(x) - g(x) = x3 + x + 1
f(x) * g(x= x5 + 3x2 - 2x 2
Figures 4.3a through 4.3c show the manual calculations. We comment on division subsequently.
Polynomial Arithmetic with Coefficients in Zp
Let us now consider polynomials in which the coefficients are elements of some field F; we refer to this as a polynomial over the field F. In that case, it is easy to show that the set of such polynomials is a ring, referred to as a polynomial ring. That is, if we consider each distinct polynomial to be an element of the set, then that set is a ring.
When polynomial arithmetic is performed on polynomials over a field, then division is possible. Note that this does not mean that exact division is possible. Let us clarify this distinction. Within a field, given two elements a and b, the quotient a/bis also an element of the field. However, given a ring R that is not a field, in
general, division will result in both a quotient and a remainder; this is not exact division.
Consider the division 5/3 within a set S. If S is the set of rational numbers, which is a field, then the result is simply expressed as 5/3 and is an element of S. Now suppose that S is the field Z7. In this case, we calculate (using Table 4.5c)
5/3 = (5 * 3 - 1) mod 7 = (5 * 5) mod 7 = 4
which is an exact solution. Finally, suppose that S is the set of integers, which is a ring but not a field. Then 5/3 produces a quotient of 1 and a remainder of 2:
5/3 = 1 + 2/3
5 = 1 * 3 + 2
Thus, division is not exact over the set of integers
Now, if we attempt to perform polynomial division over a coefficient set that is not a field, we find that division is not always defined.
If the coefficient set is the integers, then (5x2)/(3x) does not have a solution, because it would require a coefficient with a value of 5/3, which is not in the coef- ficient set. Suppose that we perform the same polynomial division over Z7. Then we have (5x2)/(3x) = 4x, which is a valid polynomial over Z7.
However, as we demonstrate presently, even if the coefficient set is a field, poly- nomial division is not necessarily exact. In general, division will produce a quotient and a remainder. We can restate the division algorithm of Equation (4.1) for polynomials over a field as follows. Given polynomials f(x) of degree n and g(x) of degree (m), (n Ú m), if we divide f(x) by g(x), we get a quotient q(x) and a remainder r(x) that obey the relationship
With the understanding that remainders are allowed, we can say that polynomial division is possible if the coefficient set is a field.
In an analogy to integer arithmetic, we can write f(x) mod g(x) for the remain- der r(x) in Equation (4.10). That is, r(x) = f(x) mod g(x). If there is no remainder [i.e., r(x) = 0], then we can say g(x) divides f(x), written as g(x) ƒ f(x). Equivalently, we can say that g(x) is a factor of f(x) or g(x) is a divisor of f(x).
For the preceding example [f(x) = x3 + x2 + 2 and g(x) = x2 - x + 1], f(x)/g(x) produces a quotient of q(x) = x + 2 and a remainder r(x) = x, as shown in Figure 4.3d. This is easily verified by noting that
q(x)g(x) + r(x) = (x + 2)(x2 - x + 1) + x = (x3 + x2 - x + 2) + x
= x3 + x2 + 2 = f(x)
For our purposes, polynomials over GF(2) are of most interest. Recall from Section 4.5 that in GF(2), addition is equivalent to the XOR operation, and multi- plication is equivalent to the logical AND operation. Further, addition and subtraction are equivalent mod 2: 1 + 1 = 1 - 1 = 0; 1 + 0 = 1 - 0 = 1; 0 + 1 = 0 - 1 = 1.
Figure 4.4 shows an example of polynomial arithmetic over GF(2). For f(x) = (x7 + x5 + x4 + x3 + x + 1) and g(x) = (x3 + x + 1), the figure shows f(x) + g(x); f(x) - g(x); f(x) * g(x); and f(x)/g(x). Note that g(x)| f(x).
A polynomial f(x) over a field F is called irreducible if and only if f(x) cannot be expressed as a product of two polynomials, both over F, and both of degree lower than that of f(x). By analogy to integers, an irreducible polynomial is also called a prime polynomial.
The polynomial9 f(xx1 over GF(2) is reducible, because
x+ (+ 1)(x+ x+ + 1).
Consider the polynomial f(x= x+ + 1. It is clear by inspection that x is not a factor of f(x). We easily show that 1 is not a factor of f(x):
Thus, f(x) has no factors of degree 1. But it is clear by inspection that if f(x) is reducible, it must have one factor of degree 2 and one factor of degree 1. Therefore, f(x) is irreducible.
Finding the Greatest Common Divisor
We can extend the analogy between polynomial arithmetic over a field and integer arithmetic by defining the greatest common divisor as follows. The polynomial c(x) is said to be the greatest common divisor of a(x) and b(x) if the following are true.
1. c(x) divides both a(x) and b(x).
2. Any divisor of a(x) and b(x) is a divisor of c(x).
An equivalent definition is the following: gcd[a(x), b(x)] is the polynomial of maximum degree that divides both a(x) and b(x).
We can adapt the Euclidean algorithm to compute the greatest common divi- sor of two polynomials. The equality in Equation (4.6) can be rewritten as the fol- lowing theorem.
gcd[a(x), b(x)] = gcd[b(x), a(x) mod b(x)] (4.11)
Equation (4.11) can be used repetitively to determine the greatest common divisor. Compare the following scheme to the definition of the Euclidean algorithm for integers.
At each iteration, we have d(x) = gcd(ri + 1(x), ri(x)) until finally d(x) = gcd(rn(x), 0) = rn(x). Thus, we can find the greatest common divisor of two integers by repetitive application of the division algorithm. This is the Euclidean algorithm for polynomials. The algorithm assumes that the degree of a(x) is greater than the degree of b(x).
Summary
We began this section with a discussion of arithmetic with ordinary polynomials. In ordinary polynomial arithmetic, the variable is not evaluated; that is, we do not plug a value in for the variable of the polynomials. Instead, arithmetic operations are per- formed on polynomials (addition, subtraction, multiplication, division) using the ordinary rules of algebra. Polynomial division is not allowed unless the coefficients are elements of a field.
Next, we discussed polynomial arithmetic in which the coefficients are elements of GF(p). In this case, polynomial addition, subtraction, multiplication, and division are allowed. However, division is not exact; that is, in general division results in a quotient and a remainder.
Finally, we showed that the Euclidean algorithm can be extended to find the greatest common divisor of two polynomials whose coefficients are elements of a field. All of the material in this section provides a foundation for the following section, in which polynomials are used to define finite fields of order pn.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Cryptography and Network Security Principles and Practice : One Symmetric Ciphers : Basic Concepts in Number Theory and Finite Fields : Polynomial Arithmetic |
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# Volume bounded by sphere and cone
Is the volume of solid bounded below by the sphere $$\rho=2 \cos(\phi)$$ and above by the cone $$z=\sqrt{x^2+y^2}$$ will be gotten by the following integration ?
in cylindrical form : $$V =\int_{0}^{2\pi}d\theta\int_{0}^{1}\int_{1+\sqrt{1-r^2}}^{r} r \,dz \, dr$$ (since the projection of their intersection will be a unit circle)
in spherical form : $$V =\int_{0}^{2\pi}d\theta\int_{\pi/4}^{\pi/2}\int_{0}^{2 \cos(\phi)}\rho^2 \sin(\phi )d\rho \, d\phi$$ But they are not giving me the same answer .. the first gives me $$-\pi$$ while the second one gives me $$\pi/3$$
If I try to get the volume by logic then it is the volume of the unit sphere when we subtract from it the volume of the cone(with unit circle base - height is 2 is the diameter of sphere) : $$4\pi/3 - 2 \pi/3= 2\pi/3$$ which is totally different from the above answers ! Note the solution manual wrote the final answer $$\pi/3$$ but I want to know what is wrong with my logic and integration in cylindrical..
In cylindrical coordinates, you need the lower half of the sphere, so when you solve $r^2+(z-1)^2=1$ for $z$, you need the negative square root. Thus \begin{align}V&=\int_0^{2\pi}\int_0^1\int_{1-\sqrt{1-r^2}}^rdz\,r\,dr\,d\theta=2\pi\int_0^1\left[r-1+\sqrt{1-r^2}\right]r\,dr\\ &=2\pi\left[\frac13r^3-\frac12r^2-\frac13(1-r^2)^{\frac32}\right]_0^1=2\pi\left[\frac13-\frac12+\frac13\right]=\frac{\pi}3\end{align} If you start with the unit sphere, then perhaps you can see from the figure that the cone in there has radius $r=1$ and height $h=1$, and you still have to take away a hemisphere on top to make a proper sno-cone, so $$V=\frac43\pi r^3-\frac13\pi r^2h-\frac12\frac43\pi r^3=\frac43\pi-\frac13\pi-\frac23\pi=\frac13\pi$$
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1. ## Checking Linear Operators
Given $u_x(x,y)-u_y(x+1,y)=xy$
$\displaystyle L(u)+L(v)=u_x(x,y)-u_y(x+1,y)+v_x(x,y)-v_y(x+1,y)=(u_x+v_x)(x,y)-(u_y+v_y)(x+1,y)=L(u+v)$
Is that how it should be written, the portion in red?
Or like this:
$\displaystyle L(u)+L(v)=u_x(x,y)-u_y(x+1,y)+v_x(x,y)-v_y(x+1,y)=u_x(x,y)+v_x(x,y)-(u_y(x+1,y)+v_y(x+1,y))=L(u+v)$
Thanks.
The color option isn't working. Which line is how it should be shown, thanks.
2. Everything you've written looks fine.
If you are worried about the expressions like:
$(u_x+v_x)(x,y)$
This is well understood to be addition of functions:
$(f + g)(z) = f(z) + g(z)$
So in this case:
$
f = u_x, g = v_x, z = (x,y)
$
which is exactly what you want.
However, you are missing the step rewriting:
$u_x+v_x = (u + v)_x$
$u_y+v_y = (u + v)_y$
This will allow the expression to match the linear operator format exactly without missing an intermediate step.
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A113582 Triangle T(n,m) read by rows: T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1. 4
1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 7, 10, 7, 1, 1, 11, 19, 19, 11, 1, 1, 16, 31, 37, 31, 16, 1, 1, 22, 46, 61, 61, 46, 22, 1, 1, 29, 64, 91, 101, 91, 64, 29, 1, 1, 37, 85, 127, 151, 151, 127, 85, 37, 1, 1, 46, 109, 169, 211, 226, 211, 169, 109, 46, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS From Paul Barry, Jan 07 2009: (Start) This triangle follows a general construction method as follows: Let a(n) be an integer sequence with a(0)=1, a(1)=1. Then T(n,k,r) := [k<=n](1+r*a(k)*a(n-k)) defines a symmetrical triangle. Row sums are n + 1 + r*Sum_{k=0..n} a(k)*a(n-k) and central coefficients are 1+r*a(n)^2. Here a(n) = C(n+1,2) and r=1. Row sums are A154322 and central coefficients are A154323. (End) LINKS G. C. Greubel, Rows n=0..100 of triangle, flattened FORMULA T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1. EXAMPLE {1}, {1, 1}, {1, 2, 1}, {1, 4, 4, 1}, {1, 7, 10, 7, 1}, {1, 11, 19, 19, 11, 1}, {1, 16, 31, 37, 31, 16, 1}, {1, 22, 46, 61, 61, 46, 22, 1}, {1, 29, 64, 91, 101, 91, 64, 29, 1}, {1, 37, 85, 127, 151, 151, 127, 85, 37, 1}, {1, 46, 109, 169, 211, 226, 211, 169, 109, 46, 1} MATHEMATICA t[n_, m_] = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]//Flatten PROG (MAGMA) /* As triangle: */ [[(n-m)*(n-m+1)*m*(m+1)/4+1: m in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 12 2016 (PARI) for(n=0, 15, for(k=0, n, print1((n-k)*(n-k+1)*k*(k+1)/4 + 1, ", "))) \\ G. C. Greubel, Aug 31 2018 CROSSREFS Sequence in context: A128562 A034368 A296157 * A347147 A295213 A118245 Adjacent sequences: A113579 A113580 A113581 * A113583 A113584 A113585 KEYWORD nonn,tabl,easy AUTHOR Roger L. Bagula, Aug 25 2008 STATUS approved
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### What math do I need to know for the Physics GRE?
Posted: Wed Aug 12, 2009 4:05 am
I was working on practice test 0177, problem 51. It is easy to solve using Malus's law. However, one step in the solution requires knowing the cosine of 45 degrees. This is easy enough with a calculator, but calculators aren't allowed on the exam (unless I am SERIOUSLY misinformed). So what do I have to do? Do I have to spend time memorizing the signs and cosines of different values so I can get questions like problem 51 correct on the exam?
While we are on the subject, another problem on test 0177 (I forget which problem) required a knowledge of the value of ln 4, or something like that. In my Physics courses, whenever I had to solve something like that I would use a calculator, which, again, isn't allowed on the exam. I can't solve it without a calculator. So what should I do? Do I have to spend time memorizing the values of different ln values for the Physics GRE? What else about math should I teach myself (or memorize)? Is there a better way to go about solving problems like 51 other than memorizing the values of different cosines? Thanks.
### Re: What math do I need to know for the Physics GRE?
Posted: Wed Aug 12, 2009 10:28 am
I would recommend you learn by heart some arithmetical values, since it is the quickest way to save time. For example, it is ok if you learn the sine and cosine of 30, 45, 60, 0 and 90 degrees by heart. Learn also by heart that SqrtP{2} = 1.41 and Sqrt{3} = 1.73. As for logarithms you need to know some basic knowledge on how to handle the logarithms with 10 as a base, how to handle sums and differences of logarithms (which turn to products and ratios of their arguments respectively), and it would be advisable to learn the values of lne = 1, ln2 = 0.69 and ln1 = 0. Learn also by heart some useful combinations of physics constants: Take for example, hc = 12.4 keV * A {A = 10^-10 m} or (kT)room = (1/40) eV. Knowing such combinations by heart can help sb manipulate the various formulas with greater convenience and more confidently! If, in the real test, you meet some other calculation like the one you mentioned then: ln4 = 2ln2 = 2* 0.7 = 1.4, roughly, and that's it. Remember also, some basic propeties of logarithms like ln(x^y) = y*lnx, x>0. Sometimes, in the real test, when the answer to one question contains strange calculations like for example the cos(15), then it is also probable to see this factor appeared as it is in the quoted answers {since it is difficult for sb to know by heart cos(15)}, but it is not always such that ... . Under different circumstances you could act like this: cos0 > cos15 > cos30 or 1 > cos15 > Sqrt{3} / 2 = 1.7 /2 = 0.85, thus you immediately know where cos(15) is confined and check the possible answers to see which fits with the result you found (even approximately).
... and as a final piece of advice: get rid of your calculator as soon as possible since it takes some time to start doing calculations "by hand". Do not defer this till the final days! When working by hand do not hesitate to make approximations (but do not overdo it, for instance do not make an approximation that would lead to a speed greater than c and so on).
### Re: What math do I need to know for the Physics GRE?
Posted: Wed Aug 12, 2009 10:37 am
These are things that you absolutely have to know, and are things that every physics major knows like the days of the week.
Memorize sin, cos, and tan of 0, 30, 45, 60, and 90 degrees and of course know their equivalents in radians.
Also memorize the basic triangles, like a 45,45,90 triange is x, x, x root(2), a 30,60,90 triange is x, 2x, x root(3)
and physics_auth is absolutely right, make sure you are not using a calculator for ANY practice problems, even if they are non-PGRE problems that are a bit harder, just practice estimating and doing back-of-the-envelope calculations.
### Re: What math do I need to know for the Physics GRE?
Posted: Sat Aug 15, 2009 7:33 am
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# Test: Integration I - Challenging
Double click on maths expressions to zoom
Question 1: Solve integral$\int \left(\frac{1}{\sqrt{1-{\mathrm{sin}}^{2}\left(3x\right)}}\right)d\left(\mathrm{sin}\left(3x\right)\right)$
$F\left(x\right)=\mathrm{sin}\left(3x\right)+C$
$F\left(x\right)=\mathrm{arcsin}\left(\mathrm{sin}\left(3x\right)\right)+C$
$F\left(x\right)=sin\left(\mathrm{arcsin}\left(3x\right)\right)+C$
$F\left(x\right)=\mathrm{arcsin}\left(\mathrm{sin}\left(3x\right)\right)$
Question 2: Find the antiderivative of the function $f\left(x\right)={\mathrm{sin}}^{2}\left(5x\right)$
$F\left(x\right)=\frac{1}{2}\left(x-\mathrm{sin}\left(10x\right)\right)+C$
$F\left(x\right)=x-\frac{1}{10}×\mathrm{sin}\left(10x\right)+C$
$F\left(x\right)=\frac{1}{2}\left(x-\frac{1}{10}×\mathrm{sin}\left(10x\right)\right)+C$
$F\left(x\right)=\frac{{\mathrm{sin}}^{3}\left(5x\right)}{3}+C$
Question 3: Find the antiderivative of the function $f\left(x\right)=\frac{1}{1+5{x}^{2}}$
$F\left(x\right)=\sqrt{5}\mathrm{arccot}\left(\sqrt{5}x\right)+C$
$F\left(x\right)=\frac{1}{\sqrt{5}}\mathrm{arccot}\left(x\right)+C$
$F\left(x\right)=5\mathrm{arccot}\left(\sqrt{5}x\right)+C$
$F\left(x\right)=\frac{1}{\sqrt{5}}\mathrm{arccot}\left(\sqrt{5}x\right)+C$
Question 4: Calculate $\underset{0}{\overset{\pi }{\int }}\left(\mathrm{cos}\left(\frac{x}{2}\right)\right)dx$
$1$
$4$
$2$
$\pi$
Question 5: Find the antiderivative of the function $f\left(x\right)=8{x}^{3}+3{x}^{2}$ that passes via point $\left(1;5\right)$
$F\left(x\right)=2{x}^{4}+{x}^{3}+C$
$F\left(x\right)=2{x}^{4}+{x}^{3}$
$F\left(x\right)=2{x}^{4}+{x}^{3}+2$
$F\left(x\right)=2{x}^{4}+{x}^{3}+5$
Question 6: Find the antiderivative of the function $f\left(x\right)={e}^{x}+3$ hat passes via point $\left(0;-7\right)$
$F\left(x\right)={e}^{x}+3x+2$
$F\left(x\right)={e}^{x}+3x+C$
$F\left(x\right)={e}^{x}+3x+8$
$F\left(x\right)={e}^{x}+3x-8$
Question 7: Calculate the area of the figure defined by $y=\sqrt{x},\text{\hspace{0.17em}}x=\frac{2}{3},\text{\hspace{0.17em}}x=\frac{8}{3}$
$2\pi$
$\frac{3\pi }{10}$
$\frac{10\pi }{3}$
$\frac{5\pi }{3}$
Question 8: Find the formula for the area of the shaded region:
$S=\underset{0}{\overset{\pi }{\int }}\left(\mathrm{sin}\left(x\right)\right)dx$
$S=\underset{0}{\overset{\pi }{\int }}\left(\mathrm{sin}\left(x\right)\right)dx-\underset{\pi }{\overset{2\pi }{\int }}\left(\mathrm{sin}\left(x\right)\right)dx$
$S=-\underset{0}{\overset{2\pi }{\int }}\left(\mathrm{sin}\left(x\right)\right)dx$
$S=\underset{0}{\overset{2\pi }{\int }}\left(\mathrm{sin}\left(x\right)\right)dx$
Question 9: Calculate the area of the figure defined by $y={x}^{2}$ and $y=3x$
$5$
$9$
$3$
$4.5$
Question 10: Calculate the area of the figure defined by $y=\frac{x}{2}$, $x=4$ and $y=0$
$6$
$7$
$2$
$4$
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A024023 a(n) = 3^n - 1. 65
0, 2, 8, 26, 80, 242, 728, 2186, 6560, 19682, 59048, 177146, 531440, 1594322, 4782968, 14348906, 43046720, 129140162, 387420488, 1162261466, 3486784400, 10460353202, 31381059608, 94143178826, 282429536480, 847288609442, 2541865828328, 7625597484986, 22876792454960 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Number of different directions along lines and hyper-diagonals in an n-dimensional cubic lattice for the attacking queens problem (A036464 in n=2, A068940 in n=3 and A068941 in n=4). The n-dimensional direction vectors have the a(n)+1 Cartesian coordinates (i,j,k,l,..) where i,j,k,l..=-1,0 or +1, excluding the zero-vector i=j=k=l=..=0. The corresponding hyper-line count is A003462. - R. J. Mathar, May 01 2006 Numbers n for which the expression 3^n/(n+1) is an integer. - Paolo P. Lava, May 29 2006 A128760(a(n)) > 0. - Reinhard Zumkeller, Mar 25 2007 Total number of sequences of length m=1,...,n with nonzero integer elements satisfying the condition Sum_{k=1..m} |n_k| <= n. See the K. A. Meissner link p. 6 (with a typo: it should be 3^([2a]-1)-1). - Wolfdieter Lang, Jan 21 2008 Draisma et al. prove that the number of lattice directions in which an n-dimensional convex body in R^n has minimal width is at most 3^n-1, with equality only for the regular cross-polytope, sharpening the 3^d-theorem of Hermann Minkowski. - Jonathan Vos Post, Jan 13 2009 [reworded from the Draisma et al. reference, Joerg Arndt, Dec 31 2017] a(n) = A024101(n)/A034472(n). - Reinhard Zumkeller, Feb 14 2009 Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x and y are disjoint and either 0) x is a proper subset of y or y is a proper subset of x, or 1) x is not a subset of y and y is not a subset of x. Then a(n) = |R|. - Ross La Haye, Mar 19 2009 Number of neighbors in Moore's neighborhood in n dimensions. - Dmitry Zaitsev, Nov 30 2015 REFERENCES Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 Omran Ahmadi, Robert Granger, An efficient deterministic test for Kloosterman sum zeros, arXiv:1104.3882 [math.NT], 2011-2012. See 1st column of Table 2 p. 9. Michael Baake, Franz Gähler, and Uwe Grimm, Examples of Substitution Systems and Their Factors, Journal of Integer Sequences, Vol. 16 (2013), #13.2.14. R. Samuel Buss, Herbrand's Theorem, University of California, Logic and Computational Complexity pp. 195-209, Lecture Notes in Computer Science, vol 960. Springer. Jan Draisma, Tyrrell B. McAllister and Benjamin Nill, Lattice width directions and Minkowski's 3^d-theorem, arXiv:0901.1375 [math.CO], Jan 10 2009. - Jonathan Vos Post, Jan 13 2009 Alessandro Farinelli, Herbrand Universe and Herbrand Base Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6. Krzysztof A. Meissner, Black hole entropy in Loop Quantum Gravity, arXiv:gr-qc/0407052, 2004. Wikipedia, Herbrand Structure Damiano Zanardini, Computational Logic, UPM European Master in Computational Logic (EMCL) School of Computer Science Technical University of Madrid, 2009-2010. Index entries for linear recurrences with constant coefficients, signature (4,-3). FORMULA a(n) = A000244(n) - 1. a(n) = 2*A003462(n). - R. J. Mathar, May 01 2006 G.f.: 2*x/(-1+x)/(-1+3*x) = 1/(-1+x)-1/(-1+3*x). - R. J. Mathar, Nov 19 2007 a(n) = Sum_{k=1..n} Sum_{m=1..k} binomial(k-1,m-1)*2^m, n>=1. a(0)=0. From the sequence combinatorics mentioned above. Twice partial sums of powers of 3. E.g.f.: e^(3*x) - e^x. - Mohammad K. Azarian, Jan 14 2009 a(n) = 3*a(n-1) + 2 (with a(0)=0). - Vincenzo Librandi, Nov 19 2010 E.g.f.: -E(0) where E(k) = 1 - 3^k/(1 - x/(x - 3^k*(k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2012 a(n) = A227048(n,A020914(n)). - Reinhard Zumkeller, Jun 30 2013 EXAMPLE From Zerinvary Lajos, Jan 14 2007: (Start) Ternary......decimal: 0...............0 2...............2 22..............8 222............26 2222...........80 22222.........242 222222........728 2222222......2186 22222222.....6560 222222222...19682 2222222222..59048 etc...........etc. (End) Sequence combinatorics: n=3: With length m=1: [1],[2],[3] each with 2 signs, with m=2: [1,1], [1,2], [2,1], each 2^2 = 4 times from choosing signs; m=3: [1,1,1] coming in 2^3 signed versions: 3*2 + 3*4 + 1*8 = 26 = a(3). The order is important, hence the M_0 multinomials A048996 enter as factors. A027902 gives the 384 divisors of a(24). - Reinhard Zumkeller, Mar 11 2010 MAPLE g:=1/(1-3*z): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-1, n=0..31); # Zerinvary Lajos, Jan 09 2009 MATHEMATICA a[0] := 0; a[n_] := a[n - 1] + 2*3^(n - 1) (* Fred Daniel Kline, Feb 09 2014 *) PROG (MAGMA) [3^n-1: n in [0..35]]; // Vincenzo Librandi, Apr 30 2011 (Haskell) a024023 = subtract 1 . a000244 -- Reinhard Zumkeller, Jun 30 2013 (PARI) a(n)=3^n-1 \\ Charles R Greathouse IV, Sep 24 2015 (PARI) vector(50, n, sum(k=0, n, 2^k*binomial(n-1, k))-1) \\ Altug Alkan, Oct 04 2015 (PARI) x='x+O('x^100); concat([0], Vec(2*x/(-1+x)/(-1+3*x))) \\ Altug Alkan, Oct 16 2015 CROSSREFS Cf. triangle A013609. Cf. A003462, A007051, A034472. Sequence in context: A124721 A279735 A103453 * A295137 A126966 A002930 Adjacent sequences: A024020 A024021 A024022 * A024024 A024025 A024026 KEYWORD nonn,easy AUTHOR STATUS approved
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# Are all angles of triangle ABC smalled than 90 degrees? 1)
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Director
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Posts: 755
Are all angles of triangle ABC smalled than 90 degrees? 1) [#permalink]
### Show Tags
04 Oct 2007, 00:54
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Are all angles of triangle ABC smalled than 90 degrees?
1) 2*AB = 3*BC = 4*AC
2) AC^2 + AB^2 = BC^2
Pls. explain.
Manager
Joined: 22 May 2007
Posts: 110
### Show Tags
04 Oct 2007, 05:11
1) SUFF
from pythagoras we know that triangle with sides 3,4 and 5 is rectangular
here we have (picking numbers) 6,4 and 3
6 > 5 -> angle is greater than 90
2) pythagoras: SUFF
not really a structured explanation but I hope this will clarify
VP
Joined: 08 Jun 2005
Posts: 1145
### Show Tags
04 Oct 2007, 06:52
statement I
2*AB = 3*BC = 4*AC
Since the sum on angles in triangle is 180 then the ratio of degrees is 180/9 = 20
2*20:3*20:4*20 = 40:60:80
sufficient
statement II
In order for all the angles of the triangle ABC be smaller than 90 degrees you cannot have a right angle. If you did then you could express the ratio of this triangle sides by the Pythagoras Theorem A^2+B^2 = C^2.
AC^2 + AB^2 = BC^2
Comply with Pythagoras Theorem. At least one angle is at 90 degrees.
sufficient
the answer is (D)
Manager
Joined: 21 Feb 2007
Posts: 80
### Show Tags
04 Oct 2007, 08:57
Statement 1 & 2 gives us diff answers,
Stat 1->40,60,80
Stat 2 ->one angle is 90
Is that possible??
Intern
Joined: 03 Oct 2007
Posts: 16
### Show Tags
04 Oct 2007, 09:12
from 1)
We know the ratios of line so we can find out the ratios of angle.
No need to calculate it.
Sufficient
from 2)
From Pythagoras Theorem one angle is 90 n others are less than 90.
Sufficient
Intern
Joined: 03 Oct 2007
Posts: 16
### Show Tags
04 Oct 2007, 09:14
abiswas wrote:
Statement 1 & 2 gives us diff answers,
Stat 1->40,60,80
Stat 2 ->one angle is 90
Is that possible??
Its possbile in GMAT,Though it rarely happens in GMAT.
04 Oct 2007, 09:14
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# Are all angles of triangle ABC smalled than 90 degrees? 1)
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NSW Syllabuses
# Mathematics K–10 - Stage 2 - Measurement and Geometry Mass
## Mass 1
### Outcomes
#### A student:
• MA2-1WM
uses appropriate terminology to describe, and symbols to represent, mathematical ideas
• MA2-3WM
checks the accuracy of a statement and explains the reasoning used
• MA2-12MG
measures, records, compares and estimates the masses of objects using kilograms and grams
### Content
• Students:
• Measure, order and compare objects using familiar metric units of mass (ACMMG061)
• recognise the need for a formal unit to measure mass
• use the kilogram as a unit to measure mass, using a pan balance
• associate kilogram measures with familiar objects, eg a standard pack of flour has a mass of 1 kg, a litre of milk has a mass of approximately 1 kg (Reasoning)
• recognise that objects with a mass of one kilogram can be a variety of shapes and sizes (Reasoning)
• record masses using the abbreviation for kilograms (kg)
• use hefting to identify objects that have a mass of 'more than', 'less than' and 'about the same as' one kilogram
• discuss strategies used to estimate mass, eg by referring to a known mass (Communicating, Problem Solving)
• compare and order two or more objects by mass measured to the nearest kilogram
• estimate the number of similar objects that have a total mass of one kilogram and check by measuring
• explain why two students may obtain different measures for the same mass (Communicating, Reasoning)
### Background Information
In Stage 2, students should appreciate that formal units allow for easier and more accurate communication of measures. Students are introduced to the kilogram and gram. They should develop an understanding of the size of these units, and use them to measure and estimate.
### Language
Students should be able to communicate using the following language: mass, more than, less than, about the same as, pan balance, (level) balance, measure, estimate, kilogram.
'Hefting' is testing the weight of an object by lifting and balancing it. Where possible, students can compare the weights of two objects by using their bodies to balance each object, eg holding one object in each hand.
As the terms 'weigh' and 'weight' are common in everyday usage, they can be accepted in student language should they arise. Weight is a force that changes with gravity, while mass remains constant.
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA2‑12MG the sub-elements (and levels) of Understanding units of measurement (UuM6-UuM7) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
## Mass 2
### Outcomes
#### A student:
• MA2-1WM
uses appropriate terminology to describe, and symbols to represent, mathematical ideas
• MA2-2WM
selects and uses appropriate mental or written strategies, or technology, to solve problems
• MA2-12MG
measures, records, compares and estimates the masses of objects using kilograms and grams
### Content
• Students:
• Use scaled instruments to measure and compare masses (ACMMG084)
• recognise the need for a formal unit smaller than the kilogram
• recognise that there are 1000 grams in one kilogram, ie 1000 grams = 1 kilogram
• use the gram as a unit to measure mass, using a scaled instrument
• associate gram measures with familiar objects, eg a standard egg has a mass of about 60 grams (Reasoning)
• record masses using the abbreviation for grams (g)
• compare two or more objects by mass measured in kilograms and grams, using a set of scales
• interpret statements, and discuss the use of kilograms and grams, on commercial packaging (Communicating, Problem Solving)
• interpret commonly used fractions of a kilogram, including $$\frac{1}{2}$$, $$\frac{1}{4}$$, $$\frac{3}{4}$$, and relate these to the number of grams
• solve problems, including those involving commonly used fractions of a kilogram (Problem Solving)
• record masses using kilograms and grams, eg 1 kg 200 g
### Background Information
Refer to background information in Mass 1.
### Language
Students should be able to communicate using the following language: mass, measure, scales, kilogram, gram.
The term 'scales', as in a set of scales, may be confusing for some students who associate it with other uses of the word 'scales', eg fish scales, scales on a map, or musical scales. These other meanings should be discussed with students.
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA2‑12MG the sub-elements (and levels) of Understanding units of measurement (UuM6-UuM8) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
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## What are the rules of sets card game?
A set consists of three cards satisfying all of these conditions:
• They all have the same number or have three different numbers.
• They all have the same shape or have three different shapes.
• They all have the same shading or have three different shadings.
• They all have the same color or have three different colors.
## How do you score a set in spades?
Scores are computed at the end of each hand and points are awarded to each team as follows:
1. If a team makes or exceeds their combined bid, they are awarded 10 points per bid.
2. If the team exceeds their combined bid, 1 point is added for each trick over their bid.
How do I get good at sets?
To find sets in Set,
1. use your naturally powerful visual pattern recognition to scan the board holistically.
2. If one feature is very common, systematically analyze only that subset.
3. continue analysis when sets are being removed and cards are being dealt.
4. anticipate useful cards or categories before they are dealt.
### How many sets are in a match?
Scoring Games, Sets and Matches There are six games in a set and two or three games in a match. Players must win a set by two games, and match by two sets.
### How many cards do you deal out in Casino?
four cards
Casino, also spelled cassino, card game for two to four players, best played with two. A 52-card deck is used. When two play, the dealer deals two cards facedown to the opponent, two cards faceup to the table, and two more facedown to himself and then repeats the process so that all have four cards.
Can you build face cards in Casino?
Face cards cannot be part of builds. Building on builds: A player may build upon another player’s build if he has a card that will total the value of the three cards in the build.
## How many sets are in the game set?
Together with the 108 unique Intercube Sets, this makes a total of 1,080 unique Sets that can be formed by the 81 cards.
## What is a 5 bag penalty in spades?
Once a team collects 5 bags, there is a “bag back” penalty: the opposing team gets an extra 50 points. Once this penalty is assessed, 5 bags are subtracted from the penalized team’s bag count. The game ends at the end of any round when a team’s score is 250 or more, or when time runs out.
How many points do you lose if you get set in spades?
Going set, no matter by how many tricks, still loses them the same 80 points while you collect your 50, but it only costs you two bags, and thus a notional 20 points.
### How do you keep score in a golf tournament?
Add up your points. When you’re finished, add up your scores from each hole for a cumulative total. Double check it. If you’re playing in a tournament, one of your competitors will be keeping your official score. You must check it and then sign your scorecard to make it official. The player with the least amount of points wins the game.
### How do you keep score in a backyard game?
In a typical backyard game, the players just keep the score themselves. Obviously, this can get a little difficult when there’s a lot of festivities around so you can look into purchasing different items to help you keep score. For example they have these suite magnetic score keepers. And now you can even download an app to help you keep score.
What are the rules of the setback card game?
Setback Card Game Rules 1 Preparing to Play. Setback is best played by two and six players, though seven- or eight-player games are possible. 2 Points in Play. Four points are possible in each hand. 3 Bidding. 4 Playing The Hand. 5 Scoring.
## How many points do you need to win a game of cards?
Place one card face-up to begin the discard pile, while remaining cards become the drawing stock. The game is won by the first player to score 500 or more points across as many hands (deals) as it takes.
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# GRE Math Basics: Distance, Rate and Time
The 4th of July weekend is upon us, and many of us will be taking road trips, or even taking a plane somewhere. To commemorate this collective movement, let’s learn the most fundamental formula when dealing with movement over time. First, let’s meet Bob…
Bob drives at an average rate of 50 mph from Berkeley to Los Angeles, a distance of 350 miles. How long does it take him to complete the trip?
(A) 4 hrs
(B) 5 hrs
(C) 7 hrs
(D) 10 hrs
(E) 12 hrs
When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time.
With the problem above, the distance between Berkeley and Los Angeles is 350 miles. So D = 350. Bob is traveling at 50 mph, so that is his rate. The question is how long will it take him to complete this trip. Therefore, we have to solve for T. Let’s set up the equation, plugging in the values for D and R:
350 = 50T.
Solving for T, we get 7. Because we are dealing with miles per hours, the 7 corresponds to hours. So T = 7. Answer (C).
Now let’s try another problem:
Charlie takes 2.5 hours to fly from Los Angeles to Mexico City, a distance of 1200 miles. What is the average speed of his plane in miles per hour?
(A) 200 mph
(B) 240 mph
(C) 410 mph
(D) 480 mph
(E) 533 mph
Setting up the equation, we get 1200 (Distance) = 2.5 (Time) x R; 1200 = 2.5R. Before solving (assuming this is on the current GRE, in which you do not have a calculator), let’s change 2.5 to 5/2, as it is much easier to do the math with fractions than with decimals. We get 1200 = 5/2R. Solving for R, we multiply both sides by the reciprocal, 2/5. This gives us R on the right-hand side of the equation, and 1200 x 2/5 on the left-hand side. 2400/5 = 480 (D).
Note you could have solved this problem back-solving, in which you put the answer choices back into the question. Or, even better, we could just plug-in 500 mph. Of course that is not one of the answer choices, but using 500 will make the math very easy. If the number is a little too high when we plug-in 500, then the answer must be (D). If it is a little low, then the answer must be (E) 533. Again, you never try to back solve with an ugly number like 533—it will take too long.
Using 500 mph, we get 1200 (Distance) = 500 (Rate) x 2.5 (Time). You can see that multiplying these two numbers gives us 1250. Meaning, we flew too fast and missed Mexico City by 50 miles. Therefore, we have to slow down the plane a little—the answer is 480.
So, this weekend, whether you are traveling abroad or just making a trip to a friend’s BBQ, you can figure out your average distance, rate, or time. Who said GRE doesn’t relate to the real world?
### 8 Responses to GRE Math Basics: Distance, Rate and Time
1. ai-ai August 16, 2016 at 8:39 pm #
can i have a step y step solution of your given problem please and a formula too
thank you
• Magoosh Test Prep Expert August 21, 2016 at 8:56 am #
Hi Ai-ai,
The formula used for these question is Distance= Rate * Time. If you have a more specific question about the step-by-step instructions provided in the blog post, please let us know! 🙂
2. Ana May 10, 2012 at 1:46 pm #
I just found a sample question in Barron’s (Example 19, pg. 161) that required knowing how to convert pounds into ounces. I am used to the metric system and don’t know by heart the conversions between pounds and ounces and the metric system, inches and feet and miles and km, Fahrenheit to Celsius and so on. I don’t remember the SAT requiring knowing such conversions. Do we have to know them for GRE?
• Chris May 10, 2012 at 2:49 pm #
Hi Ana,
Usually the question will give you the conversions. I don’t think they are testing how well you know these conversions. Nor do they want to unfairly bias those who are used to one system and not the other.
Hope that helps :).
3. Mohamed July 4, 2011 at 4:07 pm #
Thanks Chris! I’m having trouble with the ones with multiple rates… going in same/opposite directions, and early vs. late..etc can you elucidate how can I understand and train on those? Thanks alot
• Chris July 5, 2011 at 10:35 am #
Hi Mohamed, for same/opposite directions take a look at the link below:
https://magoosh.com/gre/2011/how-a-moving-train-can-be-stationary/
As for early vs. late, do you mean trains (or any other moving things) starting
at different times?
• deepak April 27, 2012 at 4:21 pm #
Hi Chris,
I really liked the Magoosh.com and also your maths videos .I am having problem in circular track questions like two persons are running on a circular track i.e. when will they meet and how many times will they meet
i know which formula to apply but the logic behind these questions i am not really care .
So plz help me with that i have also enrolled in Magoosh course Thanks a lot!
• Chris April 30, 2012 at 12:16 pm #
Hi Deepak,
The circular track questions are quite tricky :).
The key is thinking in terms of circumference, which often times means dealing with the messy pi (3.14 is not an easy number to calculate).
The other tricky part is if two entities, say race cars, are headed in opposite directions. You will have to combine their rates when figuring out how long it will take them to meet.
So if you have to cars starting at the same point and heading in opposite directions around a circular track, one going 60 km and the other 40km, you will have to combine those rates (60+40 =100). So if the track has diameter of 100 km, then the track is 314 km long, at it will take the cars 3.14 hrs to meet.
Hope that helps!
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# consider two circles, a smaller one and a larger one. If the larger one has a radius that is 3 feet larger than that of the smaller circle and the ratio of the circumferences is 2:1, what are the radii of the two circles?
Consider two circles, a smaller one and a larger one. If the larger one has a radius that is 3 feet larger than that of the smaller circle and the ratio of the circumferences is 2:1, what are the radii of the two circles?
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nezivande0u
Begin by diagramming the two circles just as it states in theproblem
Now determine the circumference of each circle
On the Right, we get 2$\pi$(r+3)
On the Left, we get 2$\pi$r
The ratio of these must be 2:1
$\frac{2}{1}=\frac{2\pi \left(r+3\right)}{2\pi r}=\frac{r+3}{r}$
now we can use the first and last fractions and use crossmultiplication to solve for r
$\left(2\right)\left(r\right)=\left(1\right)\left(r+3\right)\phantom{\rule{0ex}{0ex}}2r+r+3\phantom{\rule{0ex}{0ex}}r=3$
which is the radius of the smaller circle
r+3=6is the radius of the larger circle
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# Linear Algebra Proof on Composition of One-to-One Functions
• jrk012
In summary, to prove that the composition of one-to-one functions is also a one-to-one function, we can use a proof by contradiction. Assume that f(g(x)) is not one-to-one, then there exist a, b in set A such that f(g(a)) = f(g(b)) and a ≠ b. However, since f and g are one-to-one functions, there must exist a unique x in set B such that f(x) = f(g(a)) = f(g(b)). This contradicts the assumption that f(g(x)) is not one-to-one, thus proving that the composition of one-to-one functions is also a one-to-one function.
jrk012
## Homework Statement
Prove that the composition of one-to-one functions is also a one-to-one function.
## Homework Equations
A function is one-to-one if f(x1)=f(x2) implies x1=x2. Composition is (f*g)(x)=f(g(x)). Proof-based question.
## The Attempt at a Solution
A one-to-one function does not repeat the image. If we have two one-to-one function f(x) and g(x), then f and g do not repeat their images. Then, when then the composition, for example f(g(x)), for all x, g(x) does not repeat the image and after that applying f(x) also does not repeat the image, therefore the composition of the function is one-to-one as well.
Is this a good proof for the question?
I would suggest a proof by contradiction. Let g be a one-to-one function from set A to set B, f a one-to-one function from set B to set C. Suppose the statement were not true- that f(g(x)) is not one-to-one. Then there exist a, b, a not equal to b, in A such that f(g(a))= f(g(b)). Since f is one-to-one, there must exist a unique x in b such that f(x)= f(g(a))= f(g(b)). Can you complete this?
If in doubt, proof by contradiction!
## 1. What is Linear Algebra?
Linear Algebra is a branch of mathematics that deals with linear equations and their representations in vector spaces. It involves the study of vector operations, matrices, and linear transformations.
## 2. What is a One-to-One Function?
A one-to-one function is a type of function in which each element in the domain maps to a unique element in the range. This means that no two elements in the domain can map to the same element in the range.
## 3. What is the Composition of One-to-One Functions?
The composition of one-to-one functions is a mathematical operation that combines two one-to-one functions to create a new function. The output of one function becomes the input of the other function, resulting in a new function that is also one-to-one.
## 4. How do you prove that the composition of two One-to-One Functions is also One-to-One?
To prove that the composition of two one-to-one functions is also one-to-one, we need to show that if f and g are one-to-one functions, then the composition of f and g, denoted as f(g(x)), is also one-to-one. This can be proven by showing that for any two distinct elements in the domain of f(g(x)), their outputs will also be distinct.
## 5. How does Linear Algebra relate to the Composition of One-to-One Functions?
Linear Algebra plays a crucial role in understanding the composition of one-to-one functions. It provides the tools and concepts necessary to prove the properties of one-to-one functions and their compositions. The concepts of linear transformations and matrix operations are particularly useful in this context.
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