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labwork [276]
2 years ago
5

Referring to the Fig. in Question #34, find the midpoint of __ CD.

Mathematics
2 answers:
exis [7]2 years ago
7 0

Answer:

The answer should be (0,0.5)

Step-by-step explanation:

if it is wrong then I am sorry:'(

Alenkasestr [34]2 years ago
4 0

Answer:

(0.5,0)

Step-by-step explanation:

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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
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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

Now we have everything in order to replace into formula (1):

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)    

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