| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.95 |
| Score | 0% | 59% |
If a mayor is elected with 52% of the votes cast and 51% of a town's 10,000 voters cast a vote, how many votes did the mayor receive?
| 4,386 | |
| 2,652 | |
| 2,703 | |
| 3,111 |
If 51% of the town's 10,000 voters cast ballots the number of votes cast is:
(\( \frac{51}{100} \)) x 10,000 = \( \frac{510,000}{100} \) = 5,100
The mayor got 52% of the votes cast which is:
(\( \frac{52}{100} \)) x 5,100 = \( \frac{265,200}{100} \) = 2,652 votes.
4! = ?
4 x 3 x 2 x 1 |
|
4 x 3 |
|
3 x 2 x 1 |
|
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.
What is \( 5 \)\( \sqrt{27} \) - \( 8 \)\( \sqrt{3} \)
| 7\( \sqrt{3} \) | |
| -3\( \sqrt{81} \) | |
| 40\( \sqrt{9} \) | |
| 40\( \sqrt{27} \) |
To subtract these radicals together their radicands must be the same:
5\( \sqrt{27} \) - 8\( \sqrt{3} \)
5\( \sqrt{9 \times 3} \) - 8\( \sqrt{3} \)
5\( \sqrt{3^2 \times 3} \) - 8\( \sqrt{3} \)
(5)(3)\( \sqrt{3} \) - 8\( \sqrt{3} \)
15\( \sqrt{3} \) - 8\( \sqrt{3} \)
Now that the radicands are identical, you can subtract them:
15\( \sqrt{3} \) - 8\( \sqrt{3} \)A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 30% off." If Bob buys two shirts, each with a regular price of $15, how much money will he save?
| $0.75 | |
| $6.75 | |
| $4.50 | |
| $3.00 |
By buying two shirts, Bob will save $15 x \( \frac{30}{100} \) = \( \frac{$15 x 30}{100} \) = \( \frac{$450}{100} \) = $4.50 on the second shirt.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 60% larger than the original. By what percentage has the area of the logo increased?
| 37\(\frac{1}{2}\)% | |
| 22\(\frac{1}{2}\)% | |
| 30% | |
| 20% |
The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 60% the radius (and, consequently, the total area) increases by \( \frac{60\text{%}}{2} \) = 30%