Find the measure of angle b

A credit card company is about to send out a mailing to test the market for a new credit card. From that sample, they want to es

Sergeeva-Olga [200]

Answer:

$n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96$

And rounded up we have that n=19112

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 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:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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 $ME =\pm 0.001$ 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)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96$

And rounded up we have that n=19112

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In a 10 mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a

xenn [34]

Answer:

$0.83\text{ miles per minute}$

Step-by-step explanation:

GIVEN: In a $10$ mile race, Janet covered the first $2$ miles at a constant rate. She then speed up and rode her bike the last $8$ miles at a rate that was $0.5$ miles per minute faster. Janet's overall time would have been $2$ minutes faster had she ridden her bike the whole race at the faster pace.