| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.02 |
| Score | 0% | 60% |
If all of a roofing company's 12 workers are required to staff 3 roofing crews, how many workers need to be added during the busy season in order to send 6 complete crews out on jobs?
| 15 | |
| 17 | |
| 12 | |
| 9 |
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 3 crews so there are \( \frac{12}{3} \) = 4 workers on a crew. 6 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 6 x 4 = 24 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 24 - 12 = 12 new staff for the busy season.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 25% off." If Damon buys two shirts, each with a regular price of $38, how much money will he save?
| $5.70 | |
| $17.10 | |
| $3.80 | |
| $9.50 |
By buying two shirts, Damon will save $38 x \( \frac{25}{100} \) = \( \frac{$38 x 25}{100} \) = \( \frac{$950}{100} \) = $9.50 on the second shirt.
Solve for \( \frac{2!}{4!} \)
| \( \frac{1}{336} \) | |
| \( \frac{1}{15120} \) | |
| \( \frac{1}{12} \) | |
| \( \frac{1}{20} \) |
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:
\( \frac{2!}{4!} \)
\( \frac{2 \times 1}{4 \times 3 \times 2 \times 1} \)
\( \frac{1}{4 \times 3} \)
\( \frac{1}{12} \)
What is 3\( \sqrt{6} \) x 3\( \sqrt{6} \)?
| 6\( \sqrt{6} \) | |
| 9\( \sqrt{6} \) | |
| 54 | |
| 9\( \sqrt{12} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
3\( \sqrt{6} \) x 3\( \sqrt{6} \)
(3 x 3)\( \sqrt{6 \times 6} \)
9\( \sqrt{36} \)
Now we need to simplify the radical:
9\( \sqrt{36} \)
9\( \sqrt{6^2} \)
(9)(6)
54
Alex loaned Latoya $1,300 at an annual interest rate of 1%. If no payments are made, what is the total amount owed at the end of the first year?
| $1,404 | |
| $1,352 | |
| $1,313 | |
| $1,378 |
The yearly interest charged on this loan is the annual interest rate multiplied by the amount borrowed:
interest = annual interest rate x loan amount
i = (\( \frac{6}{100} \)) x $1,300
i = 0.01 x $1,300
No payments were made so the total amount due is the original amount + the accumulated interest:
total = $1,300 + $13