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horsena [70]
4 years ago
7

What integer describes a profit of $650?

Mathematics
2 answers:
Bas_tet [7]4 years ago
5 0

Answer:

+650

Step-by-step explanation:

The reason for this would be since it tells you it was a profit, you gained money. If you gain money it will be positive, but if you lost it it would be negative. Take it into real world- You lost 5 points posting this question but will gain 25% back if you mark me as brainliest. So make money=+650 while losing money =-650

Eduardwww [97]4 years ago
4 0

Answer:

+$650 because there is no loss of money

Step-by-step explanation:

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Step-by-step explanation:

Sample variance when large number of samples are drawn we have tend to become a normal distribution.

Variance of the sample is different since variance

Var(x) = E(x^2)-{E(x)}^2

is the formula used for finding the variance of any sample.

But sampling distribution of variance requires repeating this process with the different samples several times.

So option A is not right

B is also not correct, C is wrong

D.The sampling distribution of the mean requires multiple samples to be compared. However, in this situation you are looking for the variance, so finding the sample variance is the same as investigating the sampling distribution of the variance.

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Step-by-step explanation:

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Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc
Dima020 [189]

Answer:

a) ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

b) ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

Step-by-step explanation:

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

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)  

If solve n from equation (a) we got:  

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

Part a

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 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=\pm 1.96

If we replace the values into equation (a) for 1983 we got:

ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

Part b

Since is the same confidence level the z value it's the same.  

If we replace the values into equation (a) for 2008 we got:

ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

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Step-by-step explanation:

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