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Ksju [112]
3 years ago
7

PLEASE HELP!!! A.S.A.P!!

Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

142m

Step-by-step explanation:

This problem can be solved by simply using the pythagorean theorem, as you stated at the beginning of the problem, which is: a^{2} +b^{2} =c^{2}

You are given the <em>a</em> side and <em>b</em> side that are needed for this equation, so it's all a matter of plugging in the information you have:

110^{2} +90^{2} =c^{2}

110^{2} =12100

90^{2} =8100

12100+8100=c^{2}

20200=c^{2}

Now, because the <em>c</em> is still squared, you must take the square root of 20200 in order to get the length of just side <em>c</em>:

\sqrt{20200}≈142m

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Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47  

And rounded up we have that n=99

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".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.13 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We assume the value for \hat p =0.5 since we don't have previous info. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.13}{2.58})^2}=98.47  

And rounded up we have that n=99

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