| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.95 |
| Score | 0% | 59% |
If a mayor is elected with 75% of the votes cast and 55% of a town's 46,000 voters cast a vote, how many votes did the mayor receive?
| 20,240 | |
| 18,975 | |
| 17,457 | |
| 21,758 |
If 55% of the town's 46,000 voters cast ballots the number of votes cast is:
(\( \frac{55}{100} \)) x 46,000 = \( \frac{2,530,000}{100} \) = 25,300
The mayor got 75% of the votes cast which is:
(\( \frac{75}{100} \)) x 25,300 = \( \frac{1,897,500}{100} \) = 18,975 votes.
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 | |
| 98 m2 | |
| 32 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
4! = ?
3 x 2 x 1 |
|
5 x 4 x 3 x 2 x 1 |
|
4 x 3 |
|
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.
Solve 2 + (4 + 3) ÷ 4 x 3 - 32
| \(\frac{7}{8}\) | |
| 1\(\frac{1}{2}\) | |
| \(\frac{8}{9}\) | |
| -1\(\frac{3}{4}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
2 + (4 + 3) ÷ 4 x 3 - 32
P: 2 + (7) ÷ 4 x 3 - 32
E: 2 + 7 ÷ 4 x 3 - 9
MD: 2 + \( \frac{7}{4} \) x 3 - 9
MD: 2 + \( \frac{21}{4} \) - 9
AS: \( \frac{8}{4} \) + \( \frac{21}{4} \) - 9
AS: \( \frac{29}{4} \) - 9
AS: \( \frac{29 - 36}{4} \)
\( \frac{-7}{4} \)
-1\(\frac{3}{4}\)
What is \( \frac{-7b^5}{6b^2} \)?
| -1\(\frac{1}{6}\)b\(\frac{2}{5}\) | |
| -1\(\frac{1}{6}\)b10 | |
| -1\(\frac{1}{6}\)b3 | |
| -\(\frac{6}{7}\)b7 |
To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:
\( \frac{-7b^5}{6b^2} \)
\( \frac{-7}{6} \) b(5 - 2)
-1\(\frac{1}{6}\)b3