There are 2 5-passenger taxis available so that's 2 x 5 = 10 total seats. There are 14 people needing transportation leaving 14 - 10 = 4 who will have to find other transportation.

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 30 x \( \frac{55}{100} \) = \( \frac{55 x 30}{100} \) = \( \frac{1650}{100} \) = 16 shots

The center makes 40% 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 = \( \frac{16}{\frac{40}{100}} \) = 16 x \( \frac{100}{40} \) = \( \frac{16 x 100}{40} \) = \( \frac{1600}{40} \) = 40 shots

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

To multiply terms with radicals, multiply the coefficients and radicands separately:

4\( \sqrt{8} \) x 2\( \sqrt{8} \)
(4 x 2)\( \sqrt{8 \times 8} \)
8\( \sqrt{64} \)

Now we need to simplify the radical:

8\( \sqrt{64} \)
8\( \sqrt{8^2} \)
(8)(8)
64

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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:

\( \frac{8}{100} \) x 10 = \( \frac{8 \times 10}{100} \) = \( \frac{80}{100} \) = 0.8 errors per hour

So, in an average hour, the machine will produce 10 - 0.8 = 9.2 error free parts.

The machine ran for 24 - 5 = 19 hours yesterday so you would expect that 19 x 9.2 = 174.8 error free parts were produced yesterday.