A sample of 1500 computer chips revealed that 25% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 27% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Make the decision to reject or fail to reject the null hypothesis at the 0.10 level.

since p-value (0.0801) is  <  0.10, the result is significant

so, we reject Null hypothesis

Step-by-step explanation:

Given the data in the question;

Null Hypothesis H₀ : p = 0.27

Alternative Hypothesis H₁ : p ≠ 0.27

Test statistic

z = ( - p)/√()

we substitute

z =  (0.25 - 0.27)/√()

z = -0.02 / √( 0.1971/1500)

z = -0.02 / 0.01146  

z = -1.7452 ≈ -1.75

p-value;

for z = -1.75, the two tailed p value is 0.0801

Decision

since p-value (0.0801) is  <  0.10, the result is significant

so, we reject Null hypothesis

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.