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 5 = \( \frac{9 \times 5}{100} \) = \( \frac{45}{100} \) = 0.45 errors per hour

So, in an average hour, the machine will produce 5 - 0.45 = 4.55 error free parts.

The machine ran for 24 - 4 = 20 hours yesterday so you would expect that 20 x 4.55 = 91 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 30 meters so the equation becomes: 2w + 2h = 30.

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

2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h

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

w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5

Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m²

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

2\( \sqrt{175} \) + 3\( \sqrt{7} \)
2\( \sqrt{25 \times 7} \) + 3\( \sqrt{7} \)
2\( \sqrt{5^2 \times 7} \) + 3\( \sqrt{7} \)
(2)(5)\( \sqrt{7} \) + 3\( \sqrt{7} \)
10\( \sqrt{7} \) + 3\( \sqrt{7} \)

Now that the radicands are identical, you can add them together:

The number of students taking German or Spanish is 9 + 13 = 22. Of that group of 22, 2 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 22 - 2 = 20 who are taking at least one language. 33 - 20 = 13 students who are not taking either language.

By buying two shirts, Damon will save $20 x \( \frac{5}{100} \) = \( \frac{$20 x 5}{100} \) = \( \frac{$100}{100} \) = $1.00 on the second shirt.