A computer software vendor claims that a new version of its operating system will crash fewer than 10 times per year on average. A system administrator installs the operating system on a random sample 0f 97 computers. At the end of a year, the sample mean number of crashes is 8.9, with a standard deviation of 3.6. Does the data support the vendor’s claim? Use = 0.01.

Step-by-step explanation:

The hypothesis is written as follows

For the null hypothesis,

µd ≤ 10

For the alternative hypothesis,

µ > 10

This is a right tailed test

Since no population standard deviation is given, the distribution is a student's t.

Since n = 97

Degrees of freedom, df = n - 1 = 97 - 1 = 96

t = (x - µ)/(s/√n)

Where

x = sample mean = 8.9

µ = population mean = 10

s = samples standard deviation = 3.6

t = (8.9 - 10)/(3.6/√97) = - 3

We would determine the p value using the t test calculator. It becomes

p = 0.00172

Since alpha, 0.01 > than the p value, 0.00172, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is enough evidence that the data do not support the vendor’s claim.

answer: what do you mean

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!