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3 years ago

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Ramona is a saleswoman whose base plus commissions amounted to $52,561 last year. If her base pay was $23,960 and she made $77,3

000 in sales last year, what was her rate of commission?

2 answers:

Ksju [112]3 years ago

8 0

Answer:

37%

Step-by-step explanation:

If Ramona received an overall pay of $52,561 last year, $23,960 of which was base pay, we can first subtract these two numbers to find the amount she made from her sales commission:

$52,561 - $23,960 = $28,601

In order to find her percent commission, we can set up a proportion:

$\frac{p}{100}=\frac{28,601}{77,300}$ , where 'p' is her rate of commission

Cross-multiply and divide: 100(28601) = 77300x or 2,860,100 = 77,300x

Or x = 37%

S_A_V [24]3 years ago

8 0

Answer:

37%

Step-by-step explanation:

APEX

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Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

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Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

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$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

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$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

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$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

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$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

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$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

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