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Mamont248 [21]
4 years ago
13

What is 8.61 to the nearest tenth?

Mathematics
2 answers:
poizon [28]4 years ago
7 0
8.61 to the nearest tenth is 8.6
Anvisha [2.4K]4 years ago
3 0

8.61 would be 8.7 because you would use the 8 to round and if the number is larger then 5 then you raise the score if its lower then you lower the score

"5 or more raise the score 4 or less let it rest"

a good way to remember that

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Answer:

n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36

So the answer for this case would be n=36 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =6.995 represent the population standard deviation

n represent the sample size  

Solution to the problem

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =1.939 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

If we replace we have this:

n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36

So the answer for this case would be n=36 rounded up to the nearest integer

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