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alexira [117]
3 years ago
15

What is 84×22 in standard algorithm for fifth grade

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
7 0

Answer:

1,848

Step-by-step explanation:

   84

x  22

--------------

1848

Alekssandra [29.7K]3 years ago
5 0
84.
x22 equals 1848
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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
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Answer:

A sample of 1068 is needed.

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 is:

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

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.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

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

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

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A sample of 1068 is needed.

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Step-by-step explanation:

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