Mathematics, 12.03.2020 04:55 michaelgg9299

The heights of twelve month old boys are normally distributed with a mean of 29.8 inches and a standard deviation of 1.2 inches. About twenty-one percent of twelve month old boys are shorter than what height? Report your answer to the nearest tenths place.

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Answer from: nicolebastidas

About twenty-one percent of twelve month old boys are shorter than 28.8 inches.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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:

$\mu = 29.8, \sigma = 1.2$

About twenty-one percent of twelve month old boys are shorter than what height?

This is the value of X when Z has a pvalue of 0.21. So it is X when Z = -0.805.

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

$-0.805 = \frac{X - 29.8}{1.2}$

X - 29.8 = -0.805*1.2

X = 28.8

About twenty-one percent of twelve month old boys are shorter than 28.8 inches.

Answer from: Quest

∠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.

Answer from: Quest

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

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