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icang [17]
3 years ago
14

Iris is a saleswoman whose base pay plus commissions amounted to $69,464 last year. If her base pay was $39,710 and she made $78

,300 in sales last year, what was her rate of commission? A. 51%
B. 38%
C. 57%
D. 89%
Mathematics
1 answer:
Murljashka [212]3 years ago
6 0
Iris is a saleswoman whose base pay plus commissions amounted to $69,464 last year. If her base pay was $39,710 and she made $78,300 in sales last year, 
<span>what was her rate of commission? </span>
<span>Commissions were $29,590 against sales of $78,300 </span>
<span>Her rate of commission was 37.79%</span>
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A village fete has a children’s running race each year, run in heats of up to ten children. For each heat the first three contes
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<u>Step-by-step explanation:</u>

1)         First       and        Second         and         Third

  \dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}

2)         First       and          Second

   \dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}

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8 0
3 years ago
A student takes a driving test until it is passed.
Leya [2.2K]

Answer:

The probability that the test is taken an even number of times is 0.30.

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

P (even number of tests) = P (Passing in the 2nd attempt)

                                                  + P (Passing in the 4th attempt)

                                                       + P (Passing in the 6th attempt) ...

If a student passed in the 2nd attempt it implies that he failed in the first.

Then,  P (Passing in the 2nd attempt) = (\frac{3}{7}) \times (\frac{4}{7})

Similarly, P (Passing in the 4th attempt) = (\frac{3}{7})^{3} \times (\frac{4}{7}), since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = (\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

S=\frac{a}{1-r^{2}}

The probability is:

P (even number of tests) = (\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...

                                          =\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\  =\frac{12}{40}\\ =0.30

Thus, the probability that the test is taken an even number of times is 0.30.

7 0
3 years ago
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