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____ [38]
4 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:
Blababa [14]4 years ago
4 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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The other solution to the equation is -3.

Step-by-step explanation:

Given equation is;

5x²+bx+12=0

One solution is -4/5

We will put this solution in Eqn to get the value of b first,

5(\frac{-4}{5})^2+b(\frac{-4}{5})+12=0\\5(\frac{16}{25})+(\frac{-4b}{5})+12=0\\\frac{16}{5}-\frac{-4b}{5}=-12\\

Multiplying each term by 5

5*(\frac{16}{5})-\frac{-4b}{5}*5=-12*5\\16-4b=-60\\-4b=-60-16\\-4b=-76

Dividing both sides by -4

\frac{-4b}{-4}=\frac{-76}{-4}\\b=19

Putting b=19 in given Eqn

5x²+19x+12=0

Here,

a=5 , b=19 , c=12

Using quadratic formula,

x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\x=\frac{-(19)\±\sqrt{(19)^2-4(5)(12)}}{2(5)}\\\\x=\frac{-19\±\sqrt{361-240}}{10}\\\\x=\frac{-19\±11}{10}\\

Either,                   or,

x=\frac{-19+11}{10},\ \ \ \ \ \ x=\frac{-19-11}{10}\\x=\frac{-8}{10},\ \ \ \ \ \ \ \ \ x=\frac{-30}{10}\\x=\frac{-4}{5},\ \ \ \ \ \ \ \ \ \ x= -3

The other solution to the equation is -3.

Keywords: linear equation, quadratic formula

Learn more about quadratic formula at:

  • brainly.com/question/7128279
  • brainly.com/question/6788996

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