Suppose a batch of steel rods produced at a steel plant have a mean length of 178178 millimeters, and a standard deviation of 1212 millimeters. if 7676 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by less than 0.560.56 millimeters? round your answer to four decimal places.

To solve this problem we use the z statistic. The formula for z score is:

z = (x – u) / s

where,

x = sample value mean = 178 – 0.56 = 177.44 mm

u = total sample mean = 178 mm

s = standard deviation = 12 mm

z = (177.44 – 178) / 12

z = -0.05

from the tables P (z = -0.05) = 0.484

So the probability P is: [multiply by 2 since we account the two sides)

P = (0.5 – 0.484) * 2 = 0.0320

∠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.