Male and female high school students reported how many hours they worked each week in summer jobs. the data is represented in the following box plots:

two box plots shown. the top one is labeled males. minimum at 1, q1 at 3, median at 10.5, q3 at 14, maximum at 21. the bottom box plot is labeled females. minimum at 0, q1 at 15, median at 18, q3 at 21, no maximum shown

identify any values of data that might affect the statistical measures of spread and center.

the females worked more than the males, and the female q3 equals the top of the range.
the spread and center are skewed due to the fourth quartile missing with the females.
there is a significant outlier at the low end for the females.
the males have a high outlier, and the females have a low outlier.

50.6

step-by-step explanation:

you'll just earn points here

step-by-step explanation:

13.46

step-by-step explanation:

16.80 x .20 = 13.46

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)