What are the minimum, first quartile, median, third quartile and maximum of the data set? 45, 76, 12, 39, 87, 65, 23, 32

Rearranging the numbers given in ascending order:

12, 23, 32, 39, 45, 65, 76, 87
n = 8

Now,
Minimum = smallest number = 12

First quartile:
C = n/4 = 8/4 = 2
Since C is an even number, First quartile = (C+1)th number. That is;
First quartile = (2+1)th number = 3rd number = 32

Median:
Median is the [(n+1)/2)]th number.
(n+1)/2 = (8+1)/2 = 4.5
Averaging the two middle numbers = (39+45)/2 = 42
Therefore, median = 42

Third quartile:
Third quartile Cth number = 3n/4th number = 3*8/4 = 6th
But C is even number
Therefore, Third quartile =(C+1)th number = (6+1)th number = 7th number = 76

Maximum number = largest number = 87

The answer is 270 cars.you can do this by making a proportional relationship. for instance, 200 is the whole we are using, so on a “number line” 200 would be over 100%. now, since he sells 35% more, add 35% to 100% to get 135% on the other side of the “number line” on the bottom half, leaving the top half empty.cross multiply 200 by 135 then divide by 100 to get your answer. i hope this , if it did, , if you don’t mind, giving me brainiest! 200/100% } 200 • 135 /100 ? /135%