735 mil. Trip in 15 hours what was their average rate

Mathematics

1 answer:

nikdorinn [45]2 years ago

5 0

Answer:

`49 MPH

Step-by-step explanation:

If it took them 15 hours to go 735 miles, and you are trying to find the average rate, then divide the miles by the time (in hours), so for this specific problem, it would be 735 divided by 15 hours, and you will find the answer as 49. (MPH)

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The proprietor of a boutique in New York wanted to determine the average age of his customers. A random sample of 25 customers r

jolli1 [7]

Answer:

$28-2.064\frac{10}{\sqrt{25}}=23.872$

$28+2.064\frac{10}{\sqrt{25}}=32.128$

So on this case the 95% confidence interval would be given by (23.9;32.1)

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".

$\bar X=28$ represent the sample mean

$\mu$ population mean (variable of interest)

s=10 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that $t_{\alpha/2}=2.064$