| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.83 |
| Score | 0% | 57% |
If there were a total of 350 raffle tickets sold and you bought 31 tickets, what's the probability that you'll win the raffle?
| 10% | |
| 9% | |
| 15% | |
| 3% |
You have 31 out of the total of 350 raffle tickets sold so you have a (\( \frac{31}{350} \)) x 100 = \( \frac{31 \times 100}{350} \) = \( \frac{3100}{350} \) = 9% chance to win the raffle.
If a rectangle is twice as long as it is wide and has a perimeter of 18 meters, what is the area of the rectangle?
| 18 m2 | |
| 32 m2 | |
| 2 m2 | |
| 8 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 18 meters so the equation becomes: 2w + 2h = 18.
Putting these two equations together and solving for width (w):
2w + 2h = 18
w + h = \( \frac{18}{2} \)
w + h = 9
w = 9 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 9 - 2w
3w = 9
w = \( \frac{9}{3} \)
w = 3
Since h = 2w that makes h = (2 x 3) = 6 and the area = h x w = 3 x 6 = 18 m2
Solve 3 + (2 + 5) ÷ 4 x 2 - 42
| 1 | |
| \(\frac{1}{3}\) | |
| -9\(\frac{1}{2}\) | |
| 1\(\frac{3}{5}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
3 + (2 + 5) ÷ 4 x 2 - 42
P: 3 + (7) ÷ 4 x 2 - 42
E: 3 + 7 ÷ 4 x 2 - 16
MD: 3 + \( \frac{7}{4} \) x 2 - 16
MD: 3 + \( \frac{14}{4} \) - 16
AS: \( \frac{12}{4} \) + \( \frac{14}{4} \) - 16
AS: \( \frac{26}{4} \) - 16
AS: \( \frac{26 - 64}{4} \)
\( \frac{-38}{4} \)
-9\(\frac{1}{2}\)
| 1 | |
| 1.0 | |
| 2.8 | |
| 1.6 |
1
What is 4a3 - 9a3?
| -5a-3 | |
| 13a3 | |
| -5a3 | |
| 5a3 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:
4a3 - 9a3
(4 - 9)a3
-5a3