AMS 5 Chapter Notes - Chapter 15: Confounding, Twin, Twin Study

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peachsea-lion643

19 May 2017

School

UCSC

Department

Applied Math and Statistics

Course

AMS 5

Professor

Yonatan Katznelson

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AMS 5 Full Course Notes

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Chapter 15, #1, 2, 3, 5, 6, 8, 11. The chance it never lands six can be found by one of the following calculations. Which one, and why? (i) (1/6)10 (ii) 1 (1/6)10 (iii) (5/6)10 (iv) 1 (5/6)10 iii; since each die roll is independent from each other and you can just multiple the odds together. You may assume that the sex of a child is determined as if by drawing at random with replacement from. Another person makes up a list of all the possible committees with 5 members. One person makes up a list of all the possible committees with 2 members. Another person makes up a list of all the possible committees with 6 members. True or false: the second list is longer than the first. Find the chance that there will be exactly 2 heads among the first 5 tosses, and exactly 4 heads among the last 5 tosses.

Related Questions

PROBLEM 11: A six-sided die with sides labeled 1, 2, 3, 4, 5, and 6 is weighted so that the outcomes 1, 2, 3, 4, and 5 are all equally likely to occur. In addition, the outcome 6 is three times as likely as the outcome 1. If this die is rolled one time, what is the probability that a 6 is rolled? A) 1/2 B) 1/16 C) 3/16 D) 3/8 E) 3/7 N) none of the above

4. (21 pts) (I) (6 pts) Determine the value of a constant b for which the function g is continuous at 0. Show your work. Justify your answer! 2 x +b if x<0, X-5 g(x) = if x 20, x#4. We have to show that lim g (x) = g (0). g(0) = 0 +18 = -1; lim X0* 0 + 16 ) = lim x + 16 **0* x2 - 160 - 16- Since lim g (x) has to exist (in order for g to be continuous at x = 0), it has to be true that x-0- x0+ X- 0 lim g(x) - lim g(x) = lim g(x). Therefore, -1, and , so b = 5. so, if b = 5, lim g (x) = g(0), and therefore, g is continuous at x = 0. (II) (4 pts) Evaluate the limit. Show your work! 2. b limx- (a) lim g(x) = lim *+- 2 x + b X - 5 : 1 * +16 10 g (x) = lim x++ - Elim x2 - 16 x++ (x - 10). OR 2x+5. 2; (a) lim g (x) = lim x--00 X- (b) lim x++ 2 + 16 g(x) = lim **** x2 - 16 X-+ (III) (4 pts) Determine whether g has any horizontal asymptotes. If so, write the equation (or equations) of all horizontal asymptotes. y = 2, by part (II) (a) y = 0, by part (II) (b) (IV) (7 pts) Determine whether g has any vertical asymptotes. If so, write the equation (or equations) of all vertical asymptotes. Show your work. Justify your answer! 5 + 16 There is no vertical asymptote at x = 5, because lim g(x) = x+5 52-16' -8 + b (x) = there is no vertical asymptote at x = -4, because lim X -4 - 4 - 5 But, lim g(x) = lim + 16 x+4+ x2 - 16 lim_ (x + 16) -20 x+4+ (x - 4) 60 (x + 4) +8 (or the numerator goes to 20, and the denominator is positive and goes to o, as x + 4*) = +00 So, the function g has only one vertical asymptote, x = 4.

please solve all questions.