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2 years ago

15p+3(p-1)>3(2) exponent of 3 on two

Mathematics

1 answer:

Bas_tet [7]2 years ago

6 0

15p + 3(p-1) > 3(2^3)
15p + 3p - 3 > 3(8)
18p - 3 > 24
18p > 24 + 3
18p > 27
p > 27/18
p > 3/2 or 1 1/2

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2 years ago

Edith’s age is 1/4 of Ronald’s age. In how many years will Edith’s age be 1/3 of Ronald’s age if Edith is 20 years old?

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1 year ago

Prove that 4x-1 =11,in which the value of x=3

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2 years ago

Read 2 more answers

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p

Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

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

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is of:

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

90% confidence level, hence $\alpha = 0.9$ , z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$ , so $z = 1.645$ .

Item a:

The estimate is $\pi = 0.213 - 0.195 = 0.018$ .

The sample size is n for which M = 0.03, hence:

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

$0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}$

$\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2$

$n = 53.1$

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence $\pi = 0.05$

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

$0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2$

$n = 751.7$

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

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2 years ago