| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.31 |
| Score | 0% | 66% |
A machine in a factory has an error rate of 3 parts per 100. The machine normally runs 24 hours a day and produces 8 parts per hour. Yesterday the machine was shut down for 6 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 76 | |
| 139.7 | |
| 81.9 | |
| 115.9 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{3}{100} \) x 8 = \( \frac{3 \times 8}{100} \) = \( \frac{24}{100} \) = 0.24 errors per hour
So, in an average hour, the machine will produce 8 - 0.24 = 7.76 error free parts.
The machine ran for 24 - 6 = 18 hours yesterday so you would expect that 18 x 7.76 = 139.7 error free parts were produced yesterday.
Simplify \( \frac{36}{52} \).
| \( \frac{5}{6} \) | |
| \( \frac{2}{9} \) | |
| \( \frac{9}{13} \) | |
| \( \frac{1}{2} \) |
To simplify this fraction, first find the greatest common factor between them. The factors of 36 are [1, 2, 3, 4, 6, 9, 12, 18, 36] and the factors of 52 are [1, 2, 4, 13, 26, 52]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).
Next, divide both numerator and denominator by the GCF:
\( \frac{36}{52} \) = \( \frac{\frac{36}{4}}{\frac{52}{4}} \) = \( \frac{9}{13} \)
The __________ is the smallest positive integer that is a multiple of two or more integers.
least common multiple |
|
greatest common factor |
|
least common factor |
|
absolute value |
The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.
What is the greatest common factor of 28 and 80?
| 8 | |
| 13 | |
| 4 | |
| 20 |
The factors of 28 are [1, 2, 4, 7, 14, 28] and the factors of 80 are [1, 2, 4, 5, 8, 10, 16, 20, 40, 80]. They share 3 factors [1, 2, 4] making 4 the greatest factor 28 and 80 have in common.
Alex loaned Jennifer $700 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?
| $728 | |
| $735 | |
| $763 | |
| $749 |
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 $700
i = 0.09 x $700
No payments were made so the total amount due is the original amount + the accumulated interest:
total = $700 + $63