The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. what percent of the scores are less than 81? the scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. what percent of the scores are less than 81?

We are given the following variables:

Normal distribution

u = sample mean = 74

s = standard deviation of the samples = 7

x = sample value = 81

Since we are asked to find percent of the scores less than 81, this means we have to find the proportion using the z-score. The formula for z score is:

z = (x – u) / s

Substituting the given values:

z = (81 – 74) / 7

z = 1

Using the standard normal distribution tables for z, we find that:

P = 0.8413            at z = 1

Therefore this means that 84.13% of scores are less than 81.

x+14

step-by-step explanation:

∠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.