MATH10560 Midterm: Math10560SolutionsPE1ShortVersion

31 views12 pages

turquoisegnu128

31 Jan 2019

School

Department

Mathematics

Course

MATH10560

Professor

All

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Question 1: simplify the following expression for x x = log3 81 + log3. = log3 9 = log3 32 = 2 log3 3 = 2. Question 2: the function f (x) = x 3 + 3x + e 2x is one-to-one. By trial-and-error we determine that f 1(1) = 0 (that is f (0) = 1). f (x) = 3x 2 + 3 + 2e 2x . Hence f (f 1(1)) = f (0) = 5. Question 3: di erentiate the function f (x) = (x 2. Use logarithmic di erentiation. (take logarithm of both sides of equation, then do implicit di erentiation. ) 1) x 2+1 . x 2 1 dx = x(x 2 1)4. Make the substitution u = ln x and at x = 2e 2 have u = 2. At x = 2e, have u = 1. Question 5: which of the following expressions gives the partial fraction decomposition of the function x 2.

Related Questions

you must use matlab or octave program. others are not allowed T.T. please use only above the two program to get solution.

Problem 3: The Inverse Transformation Method A general method for simulating a RV having a continuous distribution called the inverse transformation method is based on the following result (see also Section 10.2 of the textbook and Quiz 2) Proposition 2.1. The Inverse Transformation Method). L UN U(0,1). For any continuous et distribution function F, X F-1(U) has distribution function F. (F-1(ar) is defined to equal that value y for which F(y) r.) In order to simulate X n Exp(1), for example, generate an U U(0,1) and let X log(1 -U) (see Quiz 2(c)). In fact, X log(U) is also Exp(1), since 1 U (0,1). Since X Exp(A) if X Exp(1) (why?), it follows that log(U)A has an Exp(1) distribution. This can be easily illustrated in Octave xi -log Grand (100,1)); 100 random samples from Exp (1) x2 -log (rand (100, 1))/2; 100 random samples from Exp(2) x3 -log Grand (100, 1))*2; 100 random samples from Exp(0.5) [mean (x1) mean(x2) mean (x3)] check their mean values In addition, Y X Gamma(n, A) if X Xn are i.i.d. Exp(A). Therefore, if Un are i.i.d. U(0, 1), log (Ui) 1 log Ui Gamma(n, A) (1) i-1 i-1 (a) Write an Octave code to generate 100 random samples from Gamma(10, A) by using the inverse transformation method (1). Let A 0.5, 1, 2. Plot histograms for the Gamma random samples. Octave also has a built-in gamma RV generator gamrnd (google it for reference!). Generate 100 random samples of Gamma(10, A) using gamrnd, make histograms, and compare two results. (b) As a function of U U(0, 1), create a new RV W with CDF F(w) 1 e-ti (w 0). (This is Quiz 2 (b).) Write an Octave code to generate 100 random samples of W F using the inverse transformation method (c) Consider a method for generating Weibull(a,B) RVs that have the following CDF F(z) exp(-ar 0 z oo, a Note that W in (b) is in fact Weibull(1,2). Write an Octave function myweibrnd.m that will generate n random samples for given parameters a and B. It should have the following format function CRJ myweibrnd n, alpha, beta) this function receives sample size n and parameters alpha and beta as input arguments and return R Ca vector of n random samples) (put your code here) Using your myweibrnd function, generate 1000 simulated values when (i) (a, B) (1,2) and (ii) (a,B) (2,1), respectively. Plot histograms and estimate their mean and variance

Show transcribed image text Problem 3: The Inverse Transformation Method A general method for simulating a RV having a continuous distribution called the inverse transformation method is based on the following result (see also Section 10.2 of the textbook and Quiz 2) Proposition 2.1. The Inverse Transformation Method). L UN U(0,1). For any continuous et distribution function F, X F-1(U) has distribution function F. (F-1(ar) is defined to equal that value y for which F(y) r.) In order to simulate X n Exp(1), for example, generate an U U(0,1) and let X log(1 -U) (see Quiz 2(c)). In fact, X log(U) is also Exp(1), since 1 U (0,1). Since X Exp(A) if X Exp(1) (why?), it follows that log(U)A has an Exp(1) distribution. This can be easily illustrated in Octave xi -log Grand (100,1)); 100 random samples from Exp (1) x2 -log (rand (100, 1))/2; 100 random samples from Exp(2) x3 -log Grand (100, 1))*2; 100 random samples from Exp(0.5) [mean (x1) mean(x2) mean (x3)] check their mean values In addition, Y X Gamma(n, A) if X Xn are i.i.d. Exp(A). Therefore, if Un are i.i.d. U(0, 1), log (Ui) 1 log Ui Gamma(n, A) (1) i-1 i-1 (a) Write an Octave code to generate 100 random samples from Gamma(10, A) by using the inverse transformation method (1). Let A 0.5, 1, 2. Plot histograms for the Gamma random samples. Octave also has a built-in gamma RV generator gamrnd (google it for reference!). Generate 100 random samples of Gamma(10, A) using gamrnd, make histograms, and compare two results. (b) As a function of U U(0, 1), create a new RV W with CDF F(w) 1 e-ti (w 0). (This is Quiz 2 (b).) Write an Octave code to generate 100 random samples of W F using the inverse transformation method (c) Consider a method for generating Weibull(a,B) RVs that have the following CDF F(z) exp(-ar 0 z oo, a Note that W in (b) is in fact Weibull(1,2). Write an Octave function myweibrnd.m that will generate n random samples for given parameters a and B. It should have the following format function CRJ myweibrnd n, alpha, beta) this function receives sample size n and parameters alpha and beta as input arguments and return R Ca vector of n random samples) (put your code here) Using your myweibrnd function, generate 1000 simulated values when (i) (a, B) (1,2) and (ii) (a,B) (2,1), respectively. Plot histograms and estimate their mean and variance

burgundywhale451

Q18. A size 6 dress in Country C is size 52 in Country D. A function that converts dress sizes in Country C to those in Country D is f(x) = 2(x + 20). Find a formula for the inverse of the function described.
a. f ^-1(x) = (x - 20)/2
b. f ^-1(x) = x - 20
c. f ^-1(x) = (x/2) - 20
d. f ^-1(x) = (x/2) + 20

Q19. If f(x) = 4^x and g(x) = 12, find the point of intersection of the graphs of f and g by solving f(x) = g(x). Give an exact answer.
a. (log ₄ 12, 12)
b. (log ₄ 12, 4)
c. (log ₄ 12, 0)
d. (12, 12)

Q20. Find functions f and g so that f â g = H(x) = (5 - 2x³)².
a. f(x) = 5 - 2x³ ; g(x) = x²
b. f(x) = (5 - 2x)³ ; g(x) = x²
c. f(x) = x² ; g(x) = 5 - 2x³
d. f(x) = x³ ; g(x) = (5 - 2x)²

Q29. Solve the system of equations using Cramer's Rule if it is applicable.

a. x = 9, y = 2, z = 5; (9, 2, 5)
b. x = 8, y = 4, z = 5; (8, 4, 5)
c. x = 4, y = 5, z = 4; (4, 5, 4)
d. x = 8, y = -4, z = -5; (8, -4, -5)

Q30. Solve the system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

a. x = 2, y = 3, z = -7; (2, 3, -7)
b. x = 1, y = 3, z = -4; (1, 3, -4)
c. x = 5, y = 13/2, z = -3; (5, 13/2, -3)
d. x = 1, y = -3, z = -16; (1, -3, -16)

Q33. Write the partial fraction decomposition of the rational expression (4x³ + 4x²)/(x² + 5)².
a. (4x - 4)/(x² + 5) + (-20x + 20)/(x² + 5)²
b. (4x + 4)/(x² + 5) + (-20x - 20)/(x² + 5)²
c. (4x + 4)/(x² + 5) + (20x - 20)/(x² + 5)²
d. (4x + 4)/(x² + 5) + (20x + 20)/(x² + 5)²

Q34. Solve the linear programming problem. Minimize z = 11x + 6y + 7 subject to: x = 0, y = 0, x + y = 1.
a. minimum: 7
b. minimum: 18
c. minimum: 24
d. minimum: 13

Q35. Solve the system of equations by elimination.

a. x = 10, y = 0; (10, 0)
b. x = 10, y = 10; (10, 10)
c. x = 0, y = 10; (0, 10)
d. x = 0, y = 0; (0, 0)

Q36. Perform the row operations on the given augmented matrix.

a.
b.
c.
d.

Q37. Solve the system of equations using Cramer's Rule if it is applicable.

a. x = -4, y = 3; (-4, 3)
b. x = -3, y = -4; (-3, -4)
c. x = 4, y = 3; (4, 3)
d. x = 3, y = 4; (3, 4)

Q38. Write the partial fraction decomposition of the rational expression x/(x² - 9x + 20).
a. -4/(x - 4) + 5/(x - 5)
b. -5/(x - 4) + 4/(x - 5)
c. 4/(x - 4) + -5/(x - 5)
d. -4/(x - 4) + -5/(x - 5)

Q39. Solve the system of equations.

a. x = -4, y = 4, z = -3; (-4, 4, -3)
b. x = -3, y = 4, z = -4; (-3, 4, -4)
c. x = -4, y = -3, z = 4; (-4, -3, 4)
d. inconsistent

Q40. Find the value of the determinant.

a. 224
b. 24
c. -72
d. 72