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2 years ago

11

According to the graph, how much money is spent on education?

2 answers:

Nadya [2.5K]2 years ago

5 0

Answer:

16 million

Step-by-step explanation:

The United States spends an average $15,908 per pupil on postsecondary education and $33,063 per pupil on graduate and postgraduate education. The nation's gross domestic product (GDP) grows 71.6% faster than public education spending.

romanna [79]2 years ago

4 0

16 million because they spend 33,000 per pupil

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nata0808 [166]

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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

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The margin of error:

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

For this problem, we have that:

$p = 0.1$

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 $Z = 1.96$ .

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

$0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}$

$0.04\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.04}$

$\sqrt{n} = 14.7$

$(\sqrt{n})^{2} = (14.7)^{2}$

$n = 216$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

7 0

3 years ago

Each of the past 8 days, Karen traveled 59.4 miles. How many miles did she travel in all?

madam [21]

Answer:

439.2 miles

Step-by-step explanation:

multiply the days she traveled to the amount of mile she traveled a day

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Step-by-step explanation:

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3 years ago

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Flauer [41]

Answer:

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Step-by-step explanation:

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3 years ago

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solniwko [45]

Answer:

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2 years ago