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Elanso [62]
3 years ago
7

Dayami works at Future Shop and earns $10.50/h plus 6% commission on sales. Last

Mathematics
1 answer:
skad [1K]3 years ago
7 0
<h2><u>Quick answer</u> :</h2>

= \tt\$570

<h2><u>Answer with steps</u> :</h2>

Money Dayami earns for an hour's work = $10.50

Number of hours Dayami worked last week = 40

Money Dayami would have gotten for her work :

=\tt 10.50 \times 40

=\tt  \$ \: 420

Thus, money earned by Dayami for her work = $420

Total sales Dayami made last week = $2050

Money Dayami must have earned as commission :

= 6% of total sales

= \tt6 \% \: of \: 2500

=  \tt\frac{6}{100}  \times 2500

=\tt  \frac{15000}{100}

=\tt \$150

Thus, money Dayami earned as a commission = $150

Dayami's weekly gross salary last week :

= \tt420 + 150

=\color{plum} \tt \: \$570

Thus, total salary = $570

Therefore, Dayami's gross salary last week =<u> $570</u>

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12 * (5 + 6 + 5) + (2 * 12) = 216

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$13,640

Step-by-step explanation:

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the money he had

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Let ƒ(x) = x2 + 2x and g(x) = 2x. Evaluate the composition (ƒ ∘ g)(2).
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
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Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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