| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.04 |
| Score | 0% | 61% |
If the ratio of home fans to visiting fans in a crowd is 5:1 and all 32,000 seats in a stadium are filled, how many home fans are in attendance?
| 30,400 | |
| 26,667 | |
| 31,500 | |
| 28,500 |
A ratio of 5:1 means that there are 5 home fans for every one visiting fan. So, of every 6 fans, 5 are home fans and \( \frac{5}{6} \) of every fan in the stadium is a home fan:
32,000 fans x \( \frac{5}{6} \) = \( \frac{160000}{6} \) = 26,667 fans.
A bread recipe calls for 3\(\frac{1}{8}\) cups of flour. If you only have 1\(\frac{1}{2}\) cups, how much more flour is needed?
| 1\(\frac{5}{8}\) cups | |
| \(\frac{3}{8}\) cups | |
| 2\(\frac{3}{8}\) cups | |
| 2 cups |
The amount of flour you need is (3\(\frac{1}{8}\) - 1\(\frac{1}{2}\)) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{25}{8} \) - \( \frac{12}{8} \)) cups
\( \frac{13}{8} \) cups
1\(\frac{5}{8}\) cups
Simplify \( \frac{36}{56} \).
| \( \frac{1}{4} \) | |
| \( \frac{6}{11} \) | |
| \( \frac{9}{14} \) | |
| \( \frac{4}{17} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 36 are [1, 2, 3, 4, 6, 9, 12, 18, 36] and the factors of 56 are [1, 2, 4, 7, 8, 14, 28, 56]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{36}{56} \) = \( \frac{\frac{36}{4}}{\frac{56}{4}} \) = \( \frac{9}{14} \)
Solve 2 + (3 + 5) ÷ 2 x 2 - 52
| \(\frac{2}{3}\) | |
| 1\(\frac{1}{3}\) | |
| 4 | |
| -15 |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
2 + (3 + 5) ÷ 2 x 2 - 52
P: 2 + (8) ÷ 2 x 2 - 52
E: 2 + 8 ÷ 2 x 2 - 25
MD: 2 + \( \frac{8}{2} \) x 2 - 25
MD: 2 + \( \frac{16}{2} \) - 25
AS: \( \frac{4}{2} \) + \( \frac{16}{2} \) - 25
AS: \( \frac{20}{2} \) - 25
AS: \( \frac{20 - 50}{2} \)
\( \frac{-30}{2} \)
-15
What is \( \frac{7c^5}{6c^4} \)?
| 1\(\frac{1}{6}\)c9 | |
| \(\frac{6}{7}\)c-1 | |
| 1\(\frac{1}{6}\)c1\(\frac{1}{4}\) | |
| 1\(\frac{1}{6}\)c |
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:
\( \frac{7c^5}{6c^4} \)
\( \frac{7}{6} \) c(5 - 4)
1\(\frac{1}{6}\)c