| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.01 |
| Score | 0% | 60% |
Diane scored 79% on her final exam. If each question was worth 3 points and there were 270 possible points on the exam, how many questions did Diane answer correctly?
| 86 | |
| 80 | |
| 68 | |
| 71 |
Diane scored 79% on the test meaning she earned 79% of the possible points on the test. There were 270 possible points on the test so she earned 270 x 0.79 = 213 points. Each question is worth 3 points so she got \( \frac{213}{3} \) = 71 questions right.
If there were a total of 50 raffle tickets sold and you bought 3 tickets, what's the probability that you'll win the raffle?
| 6% | |
| 11% | |
| 5% | |
| 4% |
You have 3 out of the total of 50 raffle tickets sold so you have a (\( \frac{3}{50} \)) x 100 = \( \frac{3 \times 100}{50} \) = \( \frac{300}{50} \) = 6% chance to win the raffle.
If a rectangle is twice as long as it is wide and has a perimeter of 6 meters, what is the area of the rectangle?
| 128 m2 | |
| 50 m2 | |
| 32 m2 | |
| 2 m2 |
The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 6 meters so the equation becomes: 2w + 2h = 6.
Putting these two equations together and solving for width (w):
2w + 2h = 6
w + h = \( \frac{6}{2} \)
w + h = 3
w = 3 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1
Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m2
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 25% off." If Charlie buys two shirts, each with a regular price of $25, how much money will he save?
| $6.25 | |
| $2.50 | |
| $10.00 | |
| $12.50 |
By buying two shirts, Charlie will save $25 x \( \frac{25}{100} \) = \( \frac{$25 x 25}{100} \) = \( \frac{$625}{100} \) = $6.25 on the second shirt.
Solve for \( \frac{6!}{3!} \)
| 120 | |
| 336 | |
| 72 | |
| \( \frac{1}{336} \) |
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