The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 20 x \( \frac{40}{100} \) = \( \frac{40 x 20}{100} \) = \( \frac{800}{100} \) = 8 shots

The center makes 25% of his shots so he'll have to take:

shots made = shots taken x \( \frac{\text{% made}}{100} \)
shots taken = \( \frac{\text{shots taken}}{\frac{\text{% made}}{100}} \)

to make as many shots as the guard. Plugging in values for the center gives us:

center shots taken = \( \frac{8}{\frac{25}{100}} \) = 8 x \( \frac{100}{25} \) = \( \frac{8 x 100}{25} \) = \( \frac{800}{25} \) = 32 shots

to make the same number of shots as the guard and thus score the same number of points.

In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 15 workers at the company now and that's enough to staff 5 crews so there are \( \frac{15}{5} \) = 3 workers on a crew. 10 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 10 x 3 = 30 total workers to staff the crews during the busy season. The company already employs 15 workers so they need to add 30 - 15 = 15 new staff for the busy season.

To multiply fractions, multiply the numerators together and then multiply the denominators together:

\( \frac{1}{7} \) x \( \frac{3}{6} \) = \( \frac{1 x 3}{7 x 6} \) = \( \frac{3}{42} \) = \(\frac{1}{14}\)

Average speed in miles per hour is the number of miles traveled divided by the number of hours: