| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.56 |
| Score | 0% | 71% |
How many 6-passenger vans will it take to drive all 30 members of the football team to an away game?
| 5 vans | |
| 4 vans | |
| 11 vans | |
| 14 vans |
Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:
vans = \( \frac{30}{6} \) = 5
What is the next number in this sequence: 1, 6, 11, 16, 21, __________ ?
| 35 | |
| 34 | |
| 26 | |
| 31 |
The equation for this sequence is:
an = an-1 + 5
where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:
a6 = a5 + 5
a6 = 21 + 5
a6 = 26
If \( \left|a + 3\right| \) + 1 = -8, which of these is a possible value for a?
| -7 | |
| 7 | |
| 6 | |
| 10 |
First, solve for \( \left|a + 3\right| \):
\( \left|a + 3\right| \) + 1 = -8
\( \left|a + 3\right| \) = -8 - 1
\( \left|a + 3\right| \) = -9
The value inside the absolute value brackets can be either positive or negative so (a + 3) must equal - 9 or --9 for \( \left|a + 3\right| \) to equal -9:
| a + 3 = -9 a = -9 - 3 a = -12 | a + 3 = 9 a = 9 - 3 a = 6 |
So, a = 6 or a = -12.
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 50% off." If Charlie buys two shirts, each with a regular price of $20, how much will he pay for both shirts?
| $26.00 | |
| $29.00 | |
| $10.00 | |
| $30.00 |
By buying two shirts, Charlie will save $20 x \( \frac{50}{100} \) = \( \frac{$20 x 50}{100} \) = \( \frac{$1000}{100} \) = $10.00 on the second shirt.
So, his total cost will be
$20.00 + ($20.00 - $10.00)
$20.00 + $10.00
$30.00
Solve for \( \frac{2!}{4!} \)
| 30 | |
| \( \frac{1}{20} \) | |
| \( \frac{1}{12} \) | |
| 56 |
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:
\( \frac{2!}{4!} \)
\( \frac{2 \times 1}{4 \times 3 \times 2 \times 1} \)
\( \frac{1}{4 \times 3} \)
\( \frac{1}{12} \)