Football Air Pressure During the National Football Leagueâs 2014 AFC championship game, officials measured the air pressure on of the game footballs being used by the New England Patriots. They found that the balls had an average air pressure of psi, with a standard deviation of psi. Assuming this is a representative sample of all footballs used by the Patriots in the 2014 season, perform the appropriate test to determine if the average air pressure in footballs used by the Patriots was significantly less than the allowable limit of psi. There is no extreme skewness or outliers in the data, so it is appropriate to use the -distribution. New attempt is in progress. Some of the new entries may impact the last attempt grading. Your answer is incorrect.

(a) Clearly state the null and alternative hypotheses.
(b) Give the test statistic and -value and state the conclusion of the test.
(c) Do we have evidence to conclude that the air pressure was significantly less than the allowable limit?
(d) Is it fair to assume that this sample is representative of all footballs used by the Patriots during the 2014 season?

∠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.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.