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2 years ago

Michelle is giving a haircut and a permanent to another woman. The

Mathematics

1 answer:

Solnce55 [7]2 years ago

5 0

Answer:

$77 x 15/100 = $11.55

Step-by-step explanation:

15 percent of 77 dollars is 11.55

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Thirty residents who live within a 2 mile radius of natashas school.

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1 year ago

zzz [600]

Answer:1 computer : 8 students

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Probability between 7 and 11 is 0.7799

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3 years ago

Did I get it correct?

VikaD [51]

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No they are proprotional.

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2 years ago

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Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of

maksim [4K]

Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

Out of 100 people sampled, 42 had kids.

This means that $n = 100, \pi = \frac{42}{100} = 0.42$

99% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.293$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 + 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.547$

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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2 years ago