A statistics teacher was reading through a summary of his electric usage for the previous year. Typical high temperatures for each month were included in the summary and the teacher became curious about the distribution of those temperatures. Here are the data (measured in degrees Fahrenheit). 22, 31, 64, 64, 42, 81, 79, 71, 79, 79, 52, 38

a. Find the median. Interpret this value in context.

b. Calculate the mean high temperature from last year. Show your work.

c. Which measure of center better describes a typical high temperature? Justify your answer.

64

58.5°

Step-by-step explanation:

Given the data:

22, 31, 64, 64, 42, 81, 79, 71, 79, 79, 52, 38

Rearranged data: 22, 31, 38, 42, 52, 64, 64, 71, 79, 79, 79, 81

The median temperature will be :

1/2(n + 1)th term.

n = 12 (number of samples)

0.5 * (12 + 1)th

= 0.5 (13th)

= 6.5 th term

(6th + 7th term) / 2 = (64 + 64) / 2 = 64

The middle temperature value is 80°F

The mean temperature value :

ΣX /n

= sum(22, 31, 38, 42, 52, 64, 64, 71, 79, 79, 79, 81) / 12

= 702 / 12

= 58.5° F

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder

the answer is c.

perimeters = 3√10/5

areas = 90/25

i got it down chief.