| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.98 |
| Score | 0% | 60% |
April scored 93% on her final exam. If each question was worth 2 points and there were 60 possible points on the exam, how many questions did April answer correctly?
| 16 | |
| 32 | |
| 28 | |
| 19 |
April scored 93% on the test meaning she earned 93% of the possible points on the test. There were 60 possible points on the test so she earned 60 x 0.93 = 56 points. Each question is worth 2 points so she got \( \frac{56}{2} \) = 28 questions right.
If there were a total of 350 raffle tickets sold and you bought 21 tickets, what's the probability that you'll win the raffle?
| 19% | |
| 2% | |
| 6% | |
| 16% |
You have 21 out of the total of 350 raffle tickets sold so you have a (\( \frac{21}{350} \)) x 100 = \( \frac{21 \times 100}{350} \) = \( \frac{2100}{350} \) = 6% chance to win the raffle.
What is \( \frac{7}{4} \) - \( \frac{2}{12} \)?
| 1 \( \frac{8}{12} \) | |
| 2 \( \frac{9}{15} \) | |
| 1 \( \frac{3}{12} \) | |
| 1\(\frac{7}{12}\) |
To subtract 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:
\( \frac{7 x 3}{4 x 3} \) - \( \frac{2 x 1}{12 x 1} \)
\( \frac{21}{12} \) - \( \frac{2}{12} \)
Now, because the fractions share a common denominator, you can subtract them:
\( \frac{21 - 2}{12} \) = \( \frac{19}{12} \) = 1\(\frac{7}{12}\)
What is the next number in this sequence: 1, 3, 7, 13, 21, __________ ?
| 31 | |
| 28 | |
| 34 | |
| 35 |
The equation for this sequence is:
an = an-1 + 2(n - 1)
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 + 2(6 - 1)
a6 = 21 + 2(5)
a6 = 31
How many 2\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 10 gallon tank to fill it exactly halfway?
| 2 | |
| 4 | |
| 7 | |
| 9 |
To fill a 10 gallon tank exactly halfway you'll need 5 gallons of fuel. Each fuel can holds 2\(\frac{1}{2}\) gallons so:
cans = \( \frac{5 \text{ gallons}}{2\frac{1}{2} \text{ gallons}} \) = 2