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3 years ago

11

I need a step-by step explination on how to solve this problem. I really need some help!

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer:$120

Step-by-step explanation:

85% of x= $107.27

.85x=107.27

Divide both sides by .85

X=126.20

Subtract the binder$6.20

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The square of the binomial x-4 has the form x^2-ax+16. What is the value of a?

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2 years ago

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3 years ago

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

Studentka2010 [4]

Using the z-distribution, it is found that the needed sample sizes are:

a) 242

b) 1842

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:

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

99% confidence level, hence $\alpha = 0.99$ , z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$ , so $z = 2.575$ .

Item a:

The estimate is:

$\pi = 0.223 - 0.189 = 0.034$

The sample size is <u>n for which M = 0.03</u>, then:

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

$0.03 = 2.575\sqrt{\frac{0.034(0.966)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.034(0.966)}$

$\sqrt{n} = \frac{2.575\sqrt{0.034(0.966)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.034(0.966)}}{0.03}\right)^2$

$n = 241.97$

Rounding up, a sample of 242 is needed.

Item b:

No prior estimates, hence $\pi = 0.5$ is used.

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

$0.03 = 2.575\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{2.575\sqrt{0.5(0.5)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.5(0.5)}}{0.03}\right)^2$

$n = 1841.8$

Rounding up, a sample of 1842 is needed.

For more on the z-distribution, you can check brainly.com/question/25404151

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3 years ago

What is the sum of the values in the data set? 13, 23, 21, 20, 21, 24, 18

Alborosie

24 is the answer for your problem

6 0

3 years ago