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3 years ago

8

Requirement 3. If Sierra Design wants to earn profits equal to 20% of sales revenue, how much (what total fee) should the co

mpany charge each of these two clients? Identify the formula then determine the amount Sierra Design should charge these clients. (Round your answers to the nearest whole dollar.)

1 answer:

oksano4ka [1.4K]3 years ago

6 0

Answer:

sales revenue = total costs/0.8 ; Client 1 Sales revenue = $177,550/- ; Sales revenue = $10,691/-

Explanation:

Assumptions:

As total job costs are not provided so they would be assumed(using similar question) as:

client 1 = T.C ; total job costs = $142,040.00

client 2 = G.G.Com ; total job costs = $8553.00

Given Data:

Profit = 20% of sales revenue

Solution:

As we know that

total cost = sales revenue - profit

total cost = sales revenue - 0.2*sales revenue (given: profit=0.2sales revue)

By arrangiing this equation, we get

sales revenue = total costs/0.8

Now for Client 1(T.C)

Sales revenue = total costs/ 0.8

= 142040/0.8

Sales revenue = $177,550/-

Now for Client 2 (G.G.Com)

Sales revenue = total costs/ 0.8

= 8553/ 0.8

Sales revenue = $10,691/-

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Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

$g=9.8 m/s^{2}$ is the acceleration due to gravity

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

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$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

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tekilochka [14]

Answer:

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