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3 years ago

12

Find the measure of ∠P.

1 answer:

saveliy_v [14]3 years ago

7 0

180-28-67=85. Angle P is 85 degrees

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Determine which relation is a function.

Law Incorporation [45]

Its C. because that is the only one that only one x value

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3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

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:

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

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

4 0

3 years ago

Plzzz help me yall due soon

Inessa [10]

Answer:

the answer is d ur welcome.

4 0

2 years ago

Read 2 more answers

hichkok12 [17]

Answer:

C - 19/24

Step-by-step explanation:

Take the fractions and put the denominators into a common factor

$\frac{1}{3}+ \frac{1}{3}+ \frac{1}{8}$

$\frac{8}{24}+ \frac{8}{24}+ \frac{3}{24}$

Add Numerators Across

$\frac{16}{24}+ \frac{3}{24}$

$\frac{19}{24}$

8 0

3 years ago

How can you use an algebraic expression on a coordinate plane?

nydimaria [60]

Putting it in the middle sorry if it isnt right i never was really good at these but i tried.

4 0

3 years ago