If the budget is $200 and he have 15 members then we have divide the two. 200 / 15 = $13.33 per shorts. 15x =< $200. x represents 13.33. So the solution represents the coach may spend up to $13.33 per pair of shorts. If it was even 1 cent more than $13.33 than he wouldn't have enough.So he can spend up to $13.33 or less per pair of shorts.
Answer:
15x + 9 (We combing like terms. Ex. 16x and -x)
Here is what I got for the first one. MAD=
1/(N)*(|x1-xm|+|x2-xm|+..+|xN-xm|)
=1/5(|85-83|+|83-83|+|87-83|+|90-83|+|70-83|)
=1/5(2+0+4+7+13))
=5.2
And for the second one I got, MAD=
1/(N)*(|x1-xm|+|x2-xm|+..+|xN-xm|)
=1/5(|75-76|+|74-76|+|68-76|+|83-76|+|80-76|)
=1/5(1+2+8+7+4))
=4.4
Answer:
Step-by-step explanation:
3,9,12,(15),16,18,21
mean (middle number) = 15
Q1 = 9.....the middle number of the numbers before the mean
Q3 = 18...the middle number of the numbers after the mean
interquartile range (IQR) = Q3 - Q1 = 18 - 9 = 9 <==