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3 years ago

I NEED HELP ASAP!!!!

Mathematics

1 answer:

abruzzese [7]3 years ago

8 0

Answer is 4/7

You divide the length of C'D' over the length of CD
C'D'/CD = 8/14 = 4/7

The scale factor is smaller than 1, so C'D' is shorter than CD

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Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

At x = 2

$P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}$

$P(X=2) =8.8009 \times 10^{-7}$

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At most two means at x = 0 ,1 , 2

So, $P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})$

$P(X=r) =9.9439 \times 10^{-6}$

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

$={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=0.2360$

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Answer:

Step-by-step explanation:

All votes: 10*3+18+22+15*3+20+8=143

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