9. Fun fact: 85% of adults over age 40 have children. If you randomly select 10 such adults, what is the probability that all 10 of them happen to have children? A. 85% В. 15% C. 0% D. 100% E. 19.7% F. 80.3% G. 8.5% Н. 91.5% I. None of the above 10 What are the odds that at least one of those 10 adults does not have children?

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section: Chapter Questions
Problem 14T: An unbalanced coin is weighted so that the probability of heads is 0.55. The coin is tossed ten...
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hello i need help with 9,10,11 just the multiple choice answer please and thank you i appreciate it
9. Fun fact: 85% of adults over age 40 have children. If you randomly select 10 such adults, what is
the probability that all 10 of them happen to have children?
A. 85%
В. 15%
С. 0%
D. 100%
E. 19.7%
F. 80.3%
G. 8.5%
Н. 91.5%
I. None of the above
10. What are the odds that at least one of those 10 adults does not have children?
(Same options as #9 above.)
11. If you use the standard normal distribution chart to look up the 0.6 row and the 0.03 column,
you will see the number . 7357 listed there. What does this mean?
A. P(z < .7357) = 0.63
B. P(z < 0.63) = .7357
C. P(0.03 < z < 0.6) = .7357
D. P(0.6 < z <.7357) = 0.03
E. P(z = 0.63) = .7357
Transcribed Image Text:9. Fun fact: 85% of adults over age 40 have children. If you randomly select 10 such adults, what is the probability that all 10 of them happen to have children? A. 85% В. 15% С. 0% D. 100% E. 19.7% F. 80.3% G. 8.5% Н. 91.5% I. None of the above 10. What are the odds that at least one of those 10 adults does not have children? (Same options as #9 above.) 11. If you use the standard normal distribution chart to look up the 0.6 row and the 0.03 column, you will see the number . 7357 listed there. What does this mean? A. P(z < .7357) = 0.63 B. P(z < 0.63) = .7357 C. P(0.03 < z < 0.6) = .7357 D. P(0.6 < z <.7357) = 0.03 E. P(z = 0.63) = .7357
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