Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:

vans = \( \frac{94}{13} \) = 7\(\frac{3}{13}\)

So, it will take 7 full vans and one partially full van to transport the entire team making a total of 8 vans.

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

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

2 + (5 + 4) ÷ 2 x 3 - 5²
P: 2 + (9) ÷ 2 x 3 - 5²
E: 2 + 9 ÷ 2 x 3 - 25
MD: 2 + \( \frac{9}{2} \) x 3 - 25
MD: 2 + \( \frac{27}{2} \) - 25
AS: \( \frac{4}{2} \) + \( \frac{27}{2} \) - 25
AS: \( \frac{31}{2} \) - 25
AS: \( \frac{31 - 50}{2} \)
\( \frac{-19}{2} \)
-9\(\frac{1}{2}\)

To simplify this fraction, first find the greatest common factor between them. The factors of 28 are [1, 2, 4, 7, 14, 28] and the factors of 68 are [1, 2, 4, 17, 34, 68]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{28}{68} \) = \( \frac{\frac{28}{4}}{\frac{68}{4}} \) = \( \frac{7}{17} \)

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

\( \frac{9}{100} \) x 10 = \( \frac{9 \times 10}{100} \) = \( \frac{90}{100} \) = 0.9 errors per hour

So, in an average hour, the machine will produce 10 - 0.9 = 9.1 error free parts.

The machine ran for 24 - 9 = 15 hours yesterday so you would expect that 15 x 9.1 = 136.5 error free parts were produced yesterday.