| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.01 |
| Score | 0% | 60% |
In a class of 34 students, 13 are taking German and 11 are taking Spanish. Of the students studying German or Spanish, 3 are taking both courses. How many students are not enrolled in either course?
| 29 | |
| 33 | |
| 24 | |
| 13 |
The number of students taking German or Spanish is 13 + 11 = 24. Of that group of 24, 3 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 24 - 3 = 21 who are taking at least one language. 34 - 21 = 13 students who are not taking either language.
A machine in a factory has an error rate of 7 parts per 100. The machine normally runs 24 hours a day and produces 7 parts per hour. Yesterday the machine was shut down for 7 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 110.7 | |
| 95.9 | |
| 138.3 | |
| 129.2 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{7}{100} \) x 7 = \( \frac{7 \times 7}{100} \) = \( \frac{49}{100} \) = 0.49 errors per hour
So, in an average hour, the machine will produce 7 - 0.49 = 6.51 error free parts.
The machine ran for 24 - 7 = 17 hours yesterday so you would expect that 17 x 6.51 = 110.7 error free parts were produced yesterday.
If \( \left|a + 6\right| \) + 1 = 9, which of these is a possible value for a?
| 11 | |
| -14 | |
| 17 | |
| -2 |
First, solve for \( \left|a + 6\right| \):
\( \left|a + 6\right| \) + 1 = 9
\( \left|a + 6\right| \) = 9 - 1
\( \left|a + 6\right| \) = 8
The value inside the absolute value brackets can be either positive or negative so (a + 6) must equal + 8 or -8 for \( \left|a + 6\right| \) to equal 8:
| a + 6 = 8 a = 8 - 6 a = 2 | a + 6 = -8 a = -8 - 6 a = -14 |
So, a = -14 or a = 2.
Solve 3 + (5 + 4) ÷ 5 x 2 - 22
| 2\(\frac{2}{3}\) | |
| \(\frac{7}{8}\) | |
| 1\(\frac{3}{4}\) | |
| 2\(\frac{3}{5}\) |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
3 + (5 + 4) ÷ 5 x 2 - 22
P: 3 + (9) ÷ 5 x 2 - 22
E: 3 + 9 ÷ 5 x 2 - 4
MD: 3 + \( \frac{9}{5} \) x 2 - 4
MD: 3 + \( \frac{18}{5} \) - 4
AS: \( \frac{15}{5} \) + \( \frac{18}{5} \) - 4
AS: \( \frac{33}{5} \) - 4
AS: \( \frac{33 - 20}{5} \)
\( \frac{13}{5} \)
2\(\frac{3}{5}\)
a(b + c) = ab + ac defines which of the following?
distributive property for multiplication |
|
distributive property for division |
|
commutative property for multiplication |
|
commutative property for division |
The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20.