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3 years ago

How do u calculate these numbers from least to greatest 0.02, 0, 6.98, 2 1/16, 2.2, 6.89, 2.01

Mathematics

1 answer:

levacccp [35]3 years ago

8 0

Answer:

0, 0.02, 2.01, 2 1/16, 2.2, 6.89, 6.98

Step-by-step explanation:

First thing is to turn 2 1/16 into decimal form which is 2.0625

Then look at the digit to the left of the decimal, 0 is the least, then 2, then 6

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Answer:

We need a sample size of at least 1161.

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 for the interval is:

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

For this problem, we have that:

$\pi = 0.22$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large a sample should she take?

We need a sample size of at least n.

n is found when $M = 0.02$

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

$0.02 = 1.645\sqrt{\frac{0.22*0.78}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.22*0.78}$

$\sqrt{n} = \frac{1.645\sqrt{0.22*0.78}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.22*0.78}}{0.02})^{2}$

$n = 1160.88$

Rounding up

We need a sample size of at least 1161.

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4 years ago