Agroup of 121 internet users 50-65 years of age were randomly chosen and asked to report the weekly number of hours they spend online. the purpose of the study was to determine whether the mean weekly number of hours that internet users in that age group spend online differs from the mean for internet users in general which is 12.50 hours. a 95% confidence interval for the mean weekly number of hours that internet users in that age group spend online was calculated to be (11.439 , 12.577). on the basis of this interval, at α = 0.05, what can she conclude about h0: μ = 12.50 versus ha: μ ≠ 12.50?

We support the null hypothesis.

Step-by-step explanation:

Given : The purpose of the study was to determine whether the mean weekly number of hours that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

The set of hypothesis for the given claim :-

A 95% confidence interval for the mean weekly number of hours that Internet users in that age group spend online was calculated to be (11.439 , 12.577).

Since 12.50 contained in the 95% confidence interval (11.439 , 12.577).

Therefore, we support the null hypothesis.

So the final conclusion will be, we have evidence to support the claim that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

answer: d. $20

step-by-step explanation:

just took the test

step-by-step explanation:

f(gx)=f(1/x+2)

=(1/x+2)+2

=(1+2x)/2+2

=(1+2x+4)/2

= (2x+5)/2