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sveticcg [70]
3 years ago
8

Suppose there is a 13.9 % probability that a randomly selected person aged 40 years or older is a jogger. In​ addition, there is

a 15.6 % probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. What is the probability that a randomly selected person aged 40 years or older is male and jogs question mark Would it be unusual to randomly select a person aged 40 years or older who is male and jogs question mark The probability that a randomly selected person aged 40 years or older is male and jogs is nothing ​(Round to three decimal places as​ needed.).
Mathematics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

It would be unusual to randomly select a person aged 40 years or older who is male and jogs.

Step-by-step explanation:

We have these following probabilities.

A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so P(A) = 0.13.

In​ addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. P(B/A) is the probability that the person is a male, given that he/she jogs. So P(B/A) = 0.156

The Bayes theorem states that:

P(B/A) = \frac{P(A \cap B)}{P(A)}

In which P(A \cap B) is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.

So

P(A \cap B) = P(A).P(B/A) = 0.156*0.139 = 0.217

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

A probability is unusual when it is smaller than 5%.

So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.

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Answer:

The farmer have at first <u>202</u> chickens.

Step-by-step explanation:

Given:

A farmer had twice as many chickens as ducks on his farm. After he sold 166 chickens and ducks, he had half as many chickens as ducks left.

Now, to find the chickens farmer have at first.

Let the chickens be x.

And, the ducks be y.

<em>As, the farmer had twice as many chickens as ducks on his farm.</em>

So, x=2y    ......(1)

<em>As, given the farmer after selling 166 chickens and ducks, he had half as many chickens as ducks left.</em>

According to question:

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x-166=\frac{y-29}{2}

Substituting the value of x from equation (1) we get:

2y-166=\frac{y-29}{2}

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<em>Adding both sides 332 we get:</em>

4y=y+303

<em>Subtracting both sides by </em>y<em> we get:</em>

3y=303

<em>Dividing both sides by 3 we get:</em>

y=101.

Now, to get the number of chickens substituting the value of y in equation (1):

x=2y\\\\x=2\times 101\\\\x=202.

Therefore, the farmer have at first 202 chickens.

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