Acan of soda is labeled as containing 15 fluid ounces. the quality control manager wants to verify that the filling machine is not over dash filling the cans. complete parts (a) through (d) below. (a) determine the null and alternative hypotheses that would be used to determine if the filling machine is calibrated correctly. upper h 0: mu equals 15 upper h 1: mu greater than 15 (type integers or decimals. do not round.) (b) the quality control manager obtains a sample of 78 cans and measures the contents. the sample evidence leads the manager to reject the null hypothesis. write a conclusion for this hypothesis test.

Manager has evidence that the filling machine is over dash filling the cans.

Step-by-step explanation:

Let be the average amount of fluid per can.

Given claim : A can of soda is labeled as containing 15 fluid ounces.

i.e.

The quality control manager wants to verify that the filling machine is not over dash filling the cans.

i.e.  

Since null hypothesis always contains equals sign.

So, the set of hypothesis for the given situation will be :-

Also, the quality control manager obtains a sample of 78 cans and measures the contents.

The sample evidence leads the manager to reject the null hypothesis.

That means he has evidence that the filling machine is over dash filling the cans.

∠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.

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.