Mathematics, 12.03.2020 03:37 laineyherman

Based on a poll, among adults who regret getting tattoos, 22% say that they were too young when they got their tattoos. Assume that nine adults who regret getting tattoos are randomly selected. Find the indicated probability in parts a through d. A) Find the probability that none of the selected adults say that they were too young to get tattoos. B) Find the probability that exactly one of the selected adults says that he or she was too young to get tattoos. C) Find the probability that the number of selected adults saying they were too young is 0 or 1.D) It we randomly select 9 adults. Is 1 a significantly low number who day that they were too young to get tattoos?

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Answer from: Betzabeth671

a) 10.69% probability that none of the selected adults say that they were too young to get tattoos.

b) 27.13% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

c) 37.82% probability that the number of selected adults saying they were too young is 0 or 1.

d) No, because $P(X \leq 1) 0.05$

Step-by-step explanation:

For each adult who regret getting tattoos, there are only two possible outcomes. Either they say that they were too young when then got their tattoos, or they do not say this. The probability of an adult saying hat they were too young when then got their tattoos is independent from other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

22% say that they were too young when they got their tattoos.

This means that p = 0.22

Assume that nine adults who regret getting tattoos are randomly selected.

This means that n = 9

A) Find the probability that none of the selected adults say that they were too young to get tattoos.

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.22)^{0}.(0.78)^{9} = 0.1069$

10.69% probability that none of the selected adults say that they were too young to get tattoos.

B) Find the probability that exactly one of the selected adults says that he or she was too young to get tattoos.

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{9,1}.(0.22)^{1}.(0.78)^{8} = 0.2713$

27.13% probability that exactly one of the selected adults says that he or she was too young to get tattoos.

C) Find the probability that the number of selected adults saying they were too young is 0 or 1.

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1069 + 0.2713 = 0.3782$

37.82% probability that the number of selected adults saying they were too young is 0 or 1.

D) It we randomly select 9 adults. Is 1 a significantly low number who day that they were too young to get tattoos?

x is significantly low if

$P(X \leq x) \leq 0.05$

We have that

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1069 + 0.2713 = 0.3782$

So 1 is not significantly low.

Answer from: Quest

36 is the final answer! : )

Answer from: Quest

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

Approximate the solution to this system of equations. c.which two x-values will the solution be bet

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