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2 years ago

A father is twice as old as his son and the sum of their ages is 75. How old is the father?

Mathematics

1 answer:

castortr0y [4]2 years ago

8 0

Answer: 50

Let's first of all assume the son's age as x.

So,

Father's age will be 2x i.e. twice as old as the son

Given that,

Sum of their ages = 75 years

So,

x + 2x = 75

3x = 75

x = 75/3

x = 25

Hence,

Son's age = x = 25 years

Father's age = 2x = 2 * 25 = 50 years

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A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained i

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Answer:

a) We need a sample size of at least 3109.

b) We need a sample size of at least 4145.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

(a) he uses a previous estimate of 25%?

we need a sample of size at least n.

n is found when $M = 0.02, \pi = 0.25$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.25*0.75}}{0.02})^{2}$

$n = 3108.1$

We need a sample size of at least 3109.

(b) he does not use any prior estimates?

When we do not use any prior estimate, we use $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{2.575\sqrt{0.5*0.5}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.5*0.5}}{0.02})^{2}$

$n = 4144.1$

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We need a sample size of at least 4145.

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