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

\( \frac{7}{100} \) x 5 = \( \frac{7 \times 5}{100} \) = \( \frac{35}{100} \) = 0.35 errors per hour

So, in an average hour, the machine will produce 5 - 0.35 = 4.65 error free parts.

The machine ran for 24 - 7 = 17 hours yesterday so you would expect that 17 x 4.65 = 79.1 error free parts were produced yesterday.

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

4 + (5 + 3) ÷ 3 x 5 - 4²
P: 4 + (8) ÷ 3 x 5 - 4²
E: 4 + 8 ÷ 3 x 5 - 16
MD: 4 + \( \frac{8}{3} \) x 5 - 16
MD: 4 + \( \frac{40}{3} \) - 16
AS: \( \frac{12}{3} \) + \( \frac{40}{3} \) - 16
AS: \( \frac{52}{3} \) - 16
AS: \( \frac{52 - 48}{3} \)
\( \frac{4}{3} \)
1\(\frac{1}{3}\)

The first few multiples of 2 are [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] and the first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60]. The first few multiples they share are [6, 12, 18, 24, 30] making 6 the smallest multiple 2 and 6 have in common.

A number with an exponent (b^e) consists of a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b¹ = b) and a base with an exponent of 0 equals 1 ( (b⁰ = 1).

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [30, 60, 90] making 30 the smallest multiple 6 and 10 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{8 x 5}{6 x 5} \) - \( \frac{4 x 3}{10 x 3} \)

\( \frac{40}{30} \) - \( \frac{12}{30} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{40 - 12}{30} \) = \( \frac{28}{30} \) = \(\frac{14}{15}\)