| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.33 |
| Score | 0% | 67% |
If all of a roofing company's 8 workers are required to staff 4 roofing crews, how many workers need to be added during the busy season in order to send 7 complete crews out on jobs?
| 13 | |
| 1 | |
| 15 | |
| 6 |
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 8 workers at the company now and that's enough to staff 4 crews so there are \( \frac{8}{4} \) = 2 workers on a crew. 7 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 7 x 2 = 14 total workers to staff the crews during the busy season. The company already employs 8 workers so they need to add 14 - 8 = 6 new staff for the busy season.
Which of the following is a mixed number?
\({5 \over 7} \) |
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\(1 {2 \over 5} \) |
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\({a \over 5} \) |
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\({7 \over 5} \) |
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.
If \(\left|a\right| = 7\), which of the following best describes a?
a = 7 |
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a = -7 |
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a = 7 or a = -7 |
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none of these is correct |
The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).
What is \( \frac{2}{8} \) - \( \frac{4}{10} \)?
| 1 \( \frac{1}{7} \) | |
| 1 \( \frac{6}{40} \) | |
| 2 \( \frac{3}{8} \) | |
| -\(\frac{3}{20}\) |
To subtract 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:
\( \frac{2 x 5}{8 x 5} \) - \( \frac{4 x 4}{10 x 4} \)
\( \frac{10}{40} \) - \( \frac{16}{40} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{10 - 16}{40} \) = \( \frac{-6}{40} \) = -\(\frac{3}{20}\)
Solve for \( \frac{6!}{3!} \)
| 4 | |
| \( \frac{1}{20} \) | |
| 210 | |
| 120 |
A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:
\( \frac{6!}{3!} \)
\( \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{6 \times 5 \times 4}{1} \)
\( 6 \times 5 \times 4 \)
120