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2 years ago

Which of the following World Record times from Track and Field is least

Mathematics

2 answers:

Gre4nikov [31]2 years ago

3 0

I think the answer is B. Or A.

Sidana [21]2 years ago

3 0

Answer:

 THE ANSWER SHOULD BE D OR B

Step-by-step explanation:

I DON'T KNOW

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Dovator [93]

Answer:

70x + 100

Step-by-step explanation:

Trust me on this one

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1 year ago

Joseph has 3,134 pieces of paper in his classroom.emily has 5,229 pieces of paper in her classroom.how many pieces of paper do t

AysviL [449]

2095 is the correct answer

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2 years ago

Read 2 more answers

if a distribution of raw scores were plotted and then the scores were transformed to z scores, would the shape of the distributi

nadya68 [22]

•If the raw score is transformed into a z-score, however, the value of the z-score tells exactly where the score is located relative to all the other scores in the distribution.

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2 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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:

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

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

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 $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

The minimum sample size is 2401.

4 0

2 years ago

Based on the data in the table, how many members would

bezimeni [28]

Answer:

Step-by-step explanation:

It starts going up by 6. Then it goes up by 2 more than six. 6, 8, 10. Going up 2 more than last time. 2018 would be 12 more, or 76, and 2019 would be 14 more, or 90.

5 0

2 years ago