| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.87 |
| Score | 0% | 57% |
What is the distance in miles of a trip that takes 8 hours at an average speed of 30 miles per hour?
| 315 miles | |
| 25 miles | |
| 240 miles | |
| 360 miles |
Average speed in miles per hour is the number of miles traveled divided by the number of hours:
speed = \( \frac{\text{distance}}{\text{time}} \)Solving for distance:
distance = \( \text{speed} \times \text{time} \)
distance = \( 30mph \times 8h \)
240 miles
Ezra loaned April $1,200 at an annual interest rate of 9%. If no payments are made, what is the total amount owed at the end of the first year?
| $1,284 | |
| $1,236 | |
| $1,224 | |
| $1,308 |
The yearly interest charged on this loan is the annual interest rate multiplied by the amount borrowed:
interest = annual interest rate x loan amount
i = (\( \frac{6}{100} \)) x $1,200
i = 0.09 x $1,200
No payments were made so the total amount due is the original amount + the accumulated interest:
total = $1,200 + $108If a rectangle is twice as long as it is wide and has a perimeter of 6 meters, what is the area of the rectangle?
| 2 m2 | |
| 98 m2 | |
| 18 m2 | |
| 162 m2 |
The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 6 meters so the equation becomes: 2w + 2h = 6.
Putting these two equations together and solving for width (w):
2w + 2h = 6
w + h = \( \frac{6}{2} \)
w + h = 3
w = 3 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1
Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m2
What is 5\( \sqrt{6} \) x 6\( \sqrt{5} \)?
| 11\( \sqrt{5} \) | |
| 30\( \sqrt{5} \) | |
| 30\( \sqrt{30} \) | |
| 30\( \sqrt{11} \) |
To multiply terms with radicals, multiply the coefficients and radicands separately:
5\( \sqrt{6} \) x 6\( \sqrt{5} \)
(5 x 6)\( \sqrt{6 \times 5} \)
30\( \sqrt{30} \)
On average, the center for a basketball team hits 35% of his shots while a guard on the same team hits 55% of his shots. If the guard takes 25 shots during a game, how many shots will the center have to take to score as many points as the guard assuming each shot is worth the same number of points?
| 37 | |
| 39 | |
| 50 | |
| 52 |
guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 25 x \( \frac{55}{100} \) = \( \frac{55 x 25}{100} \) = \( \frac{1375}{100} \) = 13 shots
The center makes 35% of his shots so he'll have to take:
shots made = shots taken x \( \frac{\text{% made}}{100} \)
shots taken = \( \frac{\text{shots taken}}{\frac{\text{% made}}{100}} \)
to make as many shots as the guard. Plugging in values for the center gives us:
center shots taken = \( \frac{13}{\frac{35}{100}} \) = 13 x \( \frac{100}{35} \) = \( \frac{13 x 100}{35} \) = \( \frac{1300}{35} \) = 37 shots
to make the same number of shots as the guard and thus score the same number of points.