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Fiesta28 [93]
3 years ago
9

A marketing executive is studying Internet habits of married men and women during the 8am – 10pm time period on weekends ("prime

Internet time"). Based on past data, he has determined that husbands are on the Internet 10% of the time during prime Internet time. It has also been determined that when the husband is on the Internet during prime Internet time, 40% of the time the wife is also on the Internet. When the husband is not on the Internet during prime Internet time, the wife is on the Internet 20% of the time. If the wife is on the Internet during prime Internet time, what is the probability that the husband is also on the Internet?
Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

Answer: Our required probability is 0.18.

Step-by-step explanation:

Since we have given that

Probability that husband is on the internet = 10% = 0.10

Probability that husband is not on the internet = 1-0.10 = 0.9

Probability that wife is on internet given that husband is on internet = 40% = 0.40

Probability that wife is on internet given that husband is not on internet = 20% = 0.20

Probability that wife is on internet is given by

0.1\times 0.4+0.9\times 0.2\\\\=0.04+0.18\\\\=0.22

So, Probability that the husband is also on internet given that wife is on internet is given by

\dfrac{P(\text{wife and husband both on internet)}}{P(wife\ on \ internet)}\\\\=\dfrac{0.4\times 0.1}{0.22}\\\\=\dfrac{0.04}{0.22}\\\\=0.18

Hence, our required probability is 0.18.

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The new pool is 18.84 feet larger in circumference than the old one.

<u>Explanation:</u>

Given:

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Diameter of the new pool, d = 21 feet

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We know:

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On substituting the value:

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