A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 99% confidence interval to estimate the true cost of the microwave

The 99% C.I is ( 367.703, 382.297 )

Step-by-step explanation:

Given that:

sample mean = 375

Population standard deviation = 20

Sample size n = 50

The level of significance =  1 - C.I = 1 - 0.99 = 0.01

From z tables:

The critical value = = = 2.58

The margin of error E =

The margin of error E =  

The margin of error E = 7.297

The limits of 99% C.I are expressed as:

Lower limit = = 375 - 7.297

Lower limit = 367.703

Upper limit = =  375 + 7.297

Upper limit = 382.297

The 99% C.I is =

= ( 367.703, 382.297 )

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)