A researcher is interested in finding a 90% confidence interval for the mean number of times per day that college students text. The study included 147 students who averaged 44.7 texts per

day. The standard deviation was 17.9 texts. Round answers to 3 decimal places where possible.

a. To compute the confidence interval use a tv distribution.

b. With 90% confidence the population mean number of texts per day is between

and

texts.

90% confidence the Population mean number of texts per day

(42.2561 ,47.1439)

Step-by-step explanation:

Step(i):-  

Given sample size 'n' = 147

mean of the sample size x⁻ = 44.7

standard deviation of the sample 'S' = 17.9

90% confidence the Population mean number of texts per day

Step(ii):-

Degrees of freedom

       ν=n-1=147-1=146

t₀.₁₀ =  1.6554

(44.7 - 2.4439 ,44.7 + 2.4439 )

(42.2561 ,47.1439)

Conclusion:-

90% confidence the Population mean number of texts per day

(42.2561 ,47.1439)

this answer is !

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right