e of Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE have mean 150 points and standard deviation 8.75 points. Assuming that these scores are (approximately) normally distributed, a. obtain and interpret the quartiles. b. find and interpret the 99th percentile.

a) Q1= 144.10

Median = 150

Q3=155.90

b) The 99 percentile would be:

And represent a value who accumulate 99% of the values below

Step-by-step explanation:

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

Where and

Part a

Lets begin with the first quartile:

  (a)

  (b)

We can find the quantile in the normal standard distribution and we got z=-0.674.

And we can apply the z score formula and we got:

And if we solve for a we got

The median for this case is the mean

For the third quartile we find the quantile who accumulate 0.75 of the area below and we got z=0.674 and we got:

Part b

We can find the quantile in the normal standard distribution who accumulate 0.99 of the area below and we got z=2.33.

And we can apply the z score formula and we got:

And if we solve for a we got

And represent a value who accumulate 99% of the values below

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

hello i am doing good for you that it's summer in your country or state ; )

btw in my country we just have sun and rain as (because i live in the caribbean lol)

also, thx a lot for the points : )