Janelle is training for a marathon. her weekly mileage for the last eight weeks of training is shown below.

30, 40, 41, 33, 45, 50, 34, 50

this week, she runs a total of 9 miles before suffering a minor injury and taking the rest of the week off. select the true statement about the effect of this week's mileage on janelle's weekly mileage distribution.

(a) the median weekly mileage is the same with or without the inclusion of the 9-mile week.

(b) the interquartile range of the data increases when the 9-mile week is included in the data.

(c) the average weekly mileage is the same with or without the inclusion of the 9-mile week.

(d) the median weekly mileage is higher when the 9-mile week is included in the data.

The correct answer is B.The interquartile range of the data increases when the 9-mile week is included in the data.

Step-by-step explanation:

30,33,34,40,41,45,50,50

So the median will be

40+41=81

81/2=40.5

If we add the 9 then the median will be 40 so the median becomes less so the first and the last statement are wrong.

The interquartile range will be

33+34=77

77/2=38.5

45+50=95

95/2=47.5

47.5-38.5=9

If we add the 9 then the interquartile range will be

30+40=70

70/2=35.5

50+34=84

84/2=42

84-42=42

So the interquartile range changes and the statement is right.

To calculate the average we do

30+40+41+33+45+50+34+50=323

323/8=40.375

If we add the 9 we will get

9+30+40+41+33+45+50+34+50=332

332/9=36.8

So,the average changes and the statement is wrong.

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,

the correct answer is $7.38

step-by-step explanation: