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

\( \frac{2}{100} \) x 10 = \( \frac{2 \times 10}{100} \) = \( \frac{20}{100} \) = 0.2 errors per hour

So, in an average hour, the machine will produce 10 - 0.2 = 9.8 error free parts.

The machine ran for 24 - 4 = 20 hours yesterday so you would expect that 20 x 9.8 = 196 error free parts were produced yesterday.

The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 6 meters so the equation becomes: 2w + 2h = 6.

Putting these two equations together and solving for width (w):

2w + 2h = 6
w + h = \( \frac{6}{2} \)
w + h = 3
w = 3 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1

Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m²

To subtract these radicals together their radicands must be the same:

3\( \sqrt{8} \) - 3\( \sqrt{2} \)
3\( \sqrt{4 \times 2} \) - 3\( \sqrt{2} \)
3\( \sqrt{2^2 \times 2} \) - 3\( \sqrt{2} \)
(3)(2)\( \sqrt{2} \) - 3\( \sqrt{2} \)
6\( \sqrt{2} \) - 3\( \sqrt{2} \)

Now that the radicands are identical, you can subtract them:

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.

To divide fractions, invert the second fraction and then multiply:

\( \frac{2}{5} \) ÷ \( \frac{3}{5} \) = \( \frac{2}{5} \) x \( \frac{5}{3} \)

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

\( \frac{2}{5} \) x \( \frac{5}{3} \) = \( \frac{2 x 5}{5 x 3} \) = \( \frac{10}{15} \) = \(\frac{2}{3}\)