| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.88 |
| Score | 0% | 58% |
A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 15% off." If Charlie buys two shirts, each with a regular price of $30, how much will he pay for both shirts?
| $43.50 | |
| $45.00 | |
| $33.00 | |
| $55.50 |
By buying two shirts, Charlie will save $30 x \( \frac{15}{100} \) = \( \frac{$30 x 15}{100} \) = \( \frac{$450}{100} \) = $4.50 on the second shirt.
So, his total cost will be
$30.00 + ($30.00 - $4.50)
$30.00 + $25.50
$55.50
If there were a total of 100 raffle tickets sold and you bought 9 tickets, what's the probability that you'll win the raffle?
| 5% | |
| 9% | |
| 3% | |
| 7% |
You have 9 out of the total of 100 raffle tickets sold so you have a (\( \frac{9}{100} \)) x 100 = \( \frac{9 \times 100}{100} \) = \( \frac{900}{100} \) = 9% chance to win the raffle.
Solve 2 + (5 + 2) ÷ 4 x 3 - 32
| \(\frac{2}{9}\) | |
| \(\frac{1}{3}\) | |
| -1\(\frac{3}{4}\) | |
| 2 |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
2 + (5 + 2) ÷ 4 x 3 - 32
P: 2 + (7) ÷ 4 x 3 - 32
E: 2 + 7 ÷ 4 x 3 - 9
MD: 2 + \( \frac{7}{4} \) x 3 - 9
MD: 2 + \( \frac{21}{4} \) - 9
AS: \( \frac{8}{4} \) + \( \frac{21}{4} \) - 9
AS: \( \frac{29}{4} \) - 9
AS: \( \frac{29 - 36}{4} \)
\( \frac{-7}{4} \)
-1\(\frac{3}{4}\)
In a class of 32 students, 8 are taking German and 12 are taking Spanish. Of the students studying German or Spanish, 3 are taking both courses. How many students are not enrolled in either course?
| 30 | |
| 13 | |
| 19 | |
| 15 |
The number of students taking German or Spanish is 8 + 12 = 20. Of that group of 20, 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 20 - 3 = 17 who are taking at least one language. 32 - 17 = 15 students who are not taking either language.
\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?
distributive property for division |
|
commutative property for multiplication |
|
commutative property for division |
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distributive property for multiplication |
The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).