An insurance institute conducted tests with crashes of new cars traveling at 6 mi/h. the total cost of the damages was found for a simple random sample of the tested cars and listed below. find the (a) mean, (b) median, (c) mode, and (d) midrange for the given sample data. do the different measures of center differ very much? $7 comma 542 $4 comma 853 $8 comma 989 $6 comma 294 $4 comma 215

a) Mean = $ 6378.6

b) Median = $ 6,294

c) No mode

d) Midrange = $ 6602

Step-by-step explanation:

Given data;

observations :

$7,542

$4,853

$8,989

$6,294

$4,215

total number of observations = 5

Now

a) Mean = (Sum of all the observations) / (Total number of observations )

or

Mean = ($7,542 + $4,853 + $8,989 + $6,294 + $4,215) / 5

or

Mean = $ 31893 / 5

or

Mean = $ 6378.6

b) Median

for median, arrange the data in the ascending order, we have

$4,215 , $4,853 , $6,294 , $7,542 , $8,989

the median of the 5 observations is the 3rd observation

hence, median = $ 6,294

c) Mode

mode is the highest occurring observation

here every observation is unique thus no mode

d) Midrange of the given observation

Midrange = (Highest value + lowest value) / 2

or

Midrange = ( $4,215 + $8,989 ) / 2

or

Midrange = 13204 / 2

or

Midrange = $ 6602

Yes, the different measure of the center differs much

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !