| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.75 |
| Score | 0% | 55% |
A machine in a factory has an error rate of 8 parts per 100. The machine normally runs 24 hours a day and produces 9 parts per hour. Yesterday the machine was shut down for 8 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 95.6 | |
| 92.1 | |
| 186.1 | |
| 132.5 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{8}{100} \) x 9 = \( \frac{8 \times 9}{100} \) = \( \frac{72}{100} \) = 0.72 errors per hour
So, in an average hour, the machine will produce 9 - 0.72 = 8.28 error free parts.
The machine ran for 24 - 8 = 16 hours yesterday so you would expect that 16 x 8.28 = 132.5 error free parts were produced yesterday.
If a rectangle is twice as long as it is wide and has a perimeter of 30 meters, what is the area of the rectangle?
| 50 m2 | |
| 32 m2 | |
| 8 m2 | |
| 72 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 30 meters so the equation becomes: 2w + 2h = 30.
Putting these two equations together and solving for width (w):
2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h
From the question we know that h = 2w so substituting 2w for h gives us:
w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5
Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m2
What is the least common multiple of 4 and 8?
| 16 | |
| 8 | |
| 11 | |
| 1 |
The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80]. The first few multiples they share are [8, 16, 24, 32, 40] making 8 the smallest multiple 4 and 8 have in common.
What is \( \frac{9}{5} \) + \( \frac{2}{9} \)?
| \( \frac{5}{14} \) | |
| 2\(\frac{1}{45}\) | |
| 1 \( \frac{8}{45} \) | |
| \( \frac{9}{45} \) |
To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 share.
Next, convert the fractions so each denominator equals the lowest common multiple:
\( \frac{9 x 9}{5 x 9} \) + \( \frac{2 x 5}{9 x 5} \)
\( \frac{81}{45} \) + \( \frac{10}{45} \)
Now, because the fractions share a common denominator, you can add them:
\( \frac{81 + 10}{45} \) = \( \frac{91}{45} \) = 2\(\frac{1}{45}\)
Which of the following statements about exponents is false?
b1 = b |
|
all of these are false |
|
b0 = 1 |
|
b1 = 1 |
A number with an exponent (be) consists of a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).