Let p denote the proportion of students at a particular university that use the fitness center on campus on a regular basis. For a large-sample z test of H0: p = 0.5 versus Ha: p > 0.5, find the P-value associated with each of the given values of the test statistic. (Enter your answers to four decimal places.)

(a) 1.30
(b) 0.77
(c) 1.98
(d) 2.65
(e) −0.16

a) P-value = 0.0968

b) P-value = 0.2207

c) P-value = 0.0239

d) P-value = 0.0040

e) P-value = 0.5636

Step-by-step explanation:

As the hypothesis are defined with a ">" sign, instead of an "≠", the test is right-tailed.

For this type of test, the P-value is defined as:

being z* the value for each test statistic.

The probability P is calculated from the standard normal distribution.

Then, we can calculate for each case:

(a) 1.30

(b) 0.77

(c) 1.98

(d) 2.65

(e) −0.16

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!