| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.21 |
| Score | 0% | 64% |
4! = ?
4 x 3 x 2 x 1 |
|
3 x 2 x 1 |
|
4 x 3 |
|
5 x 4 x 3 x 2 x 1 |
A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 45% off." If Frank buys two shirts, each with a regular price of $42, how much money will he save?
| $16.80 | |
| $18.90 | |
| $2.10 | |
| $6.30 |
By buying two shirts, Frank will save $42 x \( \frac{45}{100} \) = \( \frac{$42 x 45}{100} \) = \( \frac{$1890}{100} \) = $18.90 on the second shirt.
What is \( 2 \)\( \sqrt{50} \) - \( 7 \)\( \sqrt{2} \)
| 14\( \sqrt{25} \) | |
| -5\( \sqrt{100} \) | |
| 3\( \sqrt{2} \) | |
| -5\( \sqrt{-21} \) |
To subtract these radicals together their radicands must be the same:
2\( \sqrt{50} \) - 7\( \sqrt{2} \)
2\( \sqrt{25 \times 2} \) - 7\( \sqrt{2} \)
2\( \sqrt{5^2 \times 2} \) - 7\( \sqrt{2} \)
(2)(5)\( \sqrt{2} \) - 7\( \sqrt{2} \)
10\( \sqrt{2} \) - 7\( \sqrt{2} \)
Now that the radicands are identical, you can subtract them:
10\( \sqrt{2} \) - 7\( \sqrt{2} \)What is the least common multiple of 9 and 15?
| 1 | |
| 45 | |
| 41 | |
| 15 |
The first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90] and the first few multiples of 15 are [15, 30, 45, 60, 75, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 9 and 15 have in common.
If all of a roofing company's 12 workers are required to staff 4 roofing crews, how many workers need to be added during the busy season in order to send 9 complete crews out on jobs?
| 15 | |
| 19 | |
| 2 | |
| 3 |
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 12 workers at the company now and that's enough to staff 4 crews so there are \( \frac{12}{4} \) = 3 workers on a crew. 9 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 9 x 3 = 27 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 27 - 12 = 15 new staff for the busy season.