| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.41 |
| Score | 0% | 68% |
What is (b3)2?
| 2b3 | |
| b6 | |
| 3b2 | |
| b-1 |
To raise a term with an exponent to another exponent, retain the base and multiply the exponents:
(b3)2If there were a total of 250 raffle tickets sold and you bought 5 tickets, what's the probability that you'll win the raffle?
| 12% | |
| 10% | |
| 2% | |
| 8% |
You have 5 out of the total of 250 raffle tickets sold so you have a (\( \frac{5}{250} \)) x 100 = \( \frac{5 \times 100}{250} \) = \( \frac{500}{250} \) = 2% chance to win the raffle.
If \( \left|a - 1\right| \) - 7 = 9, which of these is a possible value for a?
| -15 | |
| -14 | |
| -2 | |
| 15 |
First, solve for \( \left|a - 1\right| \):
\( \left|a - 1\right| \) - 7 = 9
\( \left|a - 1\right| \) = 9 + 7
\( \left|a - 1\right| \) = 16
The value inside the absolute value brackets can be either positive or negative so (a - 1) must equal + 16 or -16 for \( \left|a - 1\right| \) to equal 16:
| a - 1 = 16 a = 16 + 1 a = 17 | a - 1 = -16 a = -16 + 1 a = -15 |
So, a = -15 or a = 17.
How many 12-passenger vans will it take to drive all 65 members of the football team to an away game?
| 6 vans | |
| 7 vans | |
| 9 vans | |
| 3 vans |
Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:
vans = \( \frac{65}{12} \) = 5\(\frac{5}{12}\)
So, it will take 5 full vans and one partially full van to transport the entire team making a total of 6 vans.
What is \( \frac{2}{4} \) + \( \frac{2}{12} \)?
| \( \frac{2}{12} \) | |
| \(\frac{2}{3}\) | |
| 1 \( \frac{5}{8} \) | |
| 1 \( \frac{4}{12} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 12 are [12, 24, 36, 48, 60, 72, 84, 96]. The first few multiples they share are [12, 24, 36, 48, 60] making 12 the smallest multiple 4 and 12 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{2 x 3}{4 x 3} \) + \( \frac{2 x 1}{12 x 1} \)
\( \frac{6}{12} \) + \( \frac{2}{12} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{6 + 2}{12} \) = \( \frac{8}{12} \) = \(\frac{2}{3}\)