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3 years ago

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ΔMNP is an isosceles triangle with MN ≈NP. If ∠ M = 3x + 1 and ∠ N = x - 11. What is the measure of ∠P?

1 answer:

s2008m [1.1K]3 years ago

4 0

Start with an equation summing all the angles in this triangle:

180 = <M + <N + <P

we are given <M and <N but not <P. But, since MN=NP, the angle <P is the same as the angle <M (isosceles, make a drawing to see). So

180 = 2<M + <N

180 = 2(3x+1) + x-11

180 = 7x - 9

x = 27

<P = 3*27+1 = 82 degrees

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Answer:

a) The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is 3.645.

Step-by-step explanation:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.7, \sigma = 0.9$

a)Find the percentage of athletes whose GPA more than 1.665.

This is 1 subtracted by the pvalue of Z when X = 1.665. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.665 - 2.7}{0.9}$

$Z = -1.15$

$Z = -1.15$ has a pvalue of 0.1251

1 - 0.1251 = 0.8749

The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is more than 85.31 percent of the athletes in the study. Compute his GPA.

His GPA is X when Z has a pvalue of 0.8531. So it is X when Z = 1.05.

$Z = \frac{X - \mu}{\sigma}$

$1.05 = \frac{X - 2.7}{0.9}$

$X - 2.7 = 1.05*0.9$

$X = 3.645$

John's GPA is 3.645.

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Answer:

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Step-by-step explanation:

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