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Alik [6]
3 years ago
11

What is the length of segment CE if AE = 15 inches and ED = 5 inches and EB = 3 inches?

Mathematics
1 answer:
Artemon [7]3 years ago
6 0

We have to determine the length of segment CE if AE = 15 inches and ED = 5 inches and EB = 3 inches.

Since ,we can observe that AB and CD are chords intersecting at point E.

Therefore, by intersecting chord theorem which states:

" When two chords intersect each other inside a circle, the products of their segments are equal"

So, AE \times EB= CE \times ED

15 \times 3= CE \times 5

45= CE \times 5

CE = \frac{45}{5}

CE = 9 inches.

So, the measure of segment CE is 9 inches.

So, Option B is the correct answer.

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Fine the value of the x in the triangle shown below.
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Answer:

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Step-by-step explanation:

The interior angles of a triangle must add to 180 degrees

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A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. The number of subjects needed to estima
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Answer:

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So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) 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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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

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

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Assuming that the deviation is known we can express the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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