In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period. Assume the population standard deviation is 1.8 hours. With a .95 probability, the margin of error is approximately .

With a .95 probability, the margin of error is = 0.392

Step-by-step explanation:

Given -

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.

Sample size ( n ) =81

Standard deviation = 1.8 hours

= 1 -.95 =.05

= = 1.96           (Using Z table)

With a .95 probability, the margin of error is

Margin of error  =  

                          =

                           =

Margin of error   = 0.392

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)