| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.14 |
| Score | 0% | 63% |
The total water usage for a city is 10,000 gallons each day. Of that total, 24% is for personal use and 46% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?
| 3,600 | |
| 12,400 | |
| 4,000 | |
| 2,200 |
46% of the water consumption is industrial use and 24% is personal use so (46% - 24%) = 22% more water is used for industrial purposes. 10,000 gallons are consumed daily so industry consumes \( \frac{22}{100} \) x 10,000 gallons = 2,200 gallons.
How many 1\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 9 gallon tank to fill it exactly halfway?
| 5 | |
| 3 | |
| 6 | |
| 3 |
To fill a 9 gallon tank exactly halfway you'll need 4\(\frac{1}{2}\) gallons of fuel. Each fuel can holds 1\(\frac{1}{2}\) gallons so:
cans = \( \frac{4\frac{1}{2} \text{ gallons}}{1\frac{1}{2} \text{ gallons}} \) = 3
Solve for \( \frac{3!}{6!} \)
| 120 | |
| \( \frac{1}{120} \) | |
| \( \frac{1}{504} \) | |
| \( \frac{1}{8} \) |
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:
\( \frac{3!}{6!} \)
\( \frac{3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6 \times 5 \times 4} \)
\( \frac{1}{120} \)
What is 3b4 - 6b4?
| 3b4 | |
| -3b4 | |
| 9b-8 | |
| 3b-4 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:
3b4 - 6b4
(3 - 6)b4
-3b4
What is -8b4 + 3b4?
| -5b16 | |
| 11b4 | |
| -5b4 | |
| -5b-8 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:
-8b4 + 3b4
(-8 + 3)b4
-5b4