| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.88 |
| Score | 0% | 58% |
If all of a roofing company's 15 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 8 complete crews out on jobs?
| 9 | |
| 17 | |
| 19 | |
| 7 |
In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 15 workers at the company now and that's enough to staff 5 crews so there are \( \frac{15}{5} \) = 3 workers on a crew. 8 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 8 x 3 = 24 total workers to staff the crews during the busy season. The company already employs 15 workers so they need to add 24 - 15 = 9 new staff for the busy season.
What is \( 4 \)\( \sqrt{18} \) - \( 6 \)\( \sqrt{2} \)
| 24\( \sqrt{2} \) | |
| -2\( \sqrt{18} \) | |
| -2\( \sqrt{-5} \) | |
| 6\( \sqrt{2} \) |
To subtract these radicals together their radicands must be the same:
4\( \sqrt{18} \) - 6\( \sqrt{2} \)
4\( \sqrt{9 \times 2} \) - 6\( \sqrt{2} \)
4\( \sqrt{3^2 \times 2} \) - 6\( \sqrt{2} \)
(4)(3)\( \sqrt{2} \) - 6\( \sqrt{2} \)
12\( \sqrt{2} \) - 6\( \sqrt{2} \)
Now that the radicands are identical, you can subtract them:
12\( \sqrt{2} \) - 6\( \sqrt{2} \)Which of the following is an improper fraction?
\({2 \over 5} \) |
|
\({a \over 5} \) |
|
\(1 {2 \over 5} \) |
|
\({7 \over 5} \) |
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.
What is \( \frac{8}{8} \) + \( \frac{5}{10} \)?
| 2 \( \frac{3}{40} \) | |
| 1\(\frac{1}{2}\) | |
| 1 \( \frac{2}{6} \) | |
| 1 \( \frac{3}{10} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [40, 80] making 40 the smallest multiple 8 and 10 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{8 x 5}{8 x 5} \) + \( \frac{5 x 4}{10 x 4} \)
\( \frac{40}{40} \) + \( \frac{20}{40} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{40 + 20}{40} \) = \( \frac{60}{40} \) = 1\(\frac{1}{2}\)
A bread recipe calls for 3\(\frac{1}{8}\) cups of flour. If you only have 1\(\frac{1}{8}\) cups, how much more flour is needed?
| 3 cups | |
| 3\(\frac{1}{8}\) cups | |
| \(\frac{3}{4}\) cups | |
| 2 cups |
The amount of flour you need is (3\(\frac{1}{8}\) - 1\(\frac{1}{8}\)) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{25}{8} \) - \( \frac{9}{8} \)) cups
\( \frac{16}{8} \) cups
2 cups