Assume that Randy’s photocopying Service charges $.10 per photocopy. If fixed costs are

$27000 a year and variable costs are $0.04 per copy.

1. How Randy can compute his breakeven point? Show the result in graph.

2. How many photocopies are required to earn $ 500 profit?

3. Identify the safety margin at breakeven point.

Answers

Answer 1

Answer:

1) 739 copies (739.7260273972603).

2) 5000 copies.

Explanation:

27000/365 = 73.97260273972603 (Their daily wins). How much times $.10 is equal to 73.97260273972603?

I used a two step equation to get it.

0.10 * x = 73.97260273972603/0.10     = /0.10         x = 739.7260273972603

They need to sell 739.7260273972603 copies a day to get 27000$ a year.

an equation (y = mx+b) can be

y = 739.7260273972603x

To earn a 500$ profit we need to do the same two-step equation but instead of 73.97260273972603 use 500.

0.10 * x = 500/0.10     = /0.10         x = 5000

5000 copies needs to be sold to get a 500$ profit.

The last question I don't understand but I hope those 2 questions helped


Related Questions

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a heating value of 141,790 kJ/kg and a density of 0.0899 kg/m3, determine the efficiency of this fuel cell.

Answers

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ([tex]\eta[/tex]), dimensionless, as:

[tex]\eta = \frac{\dot W_{out}}{\dot W_{in}}[/tex] (Eq. 1)

Where:

[tex]\dot W_{in}[/tex] - Maximum power possible from hydrogen flow, measured in kilowatts.

[tex]\dot W_{out}[/tex] - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

[tex]\dot W_{in} = \dot V\cdot \rho \cdot L_{c}[/tex] (Eq. 2)

Where:

[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.

[tex]\rho[/tex] - Density of hydrogen, measured in kilograms per cubic meter.

[tex]L_{c}[/tex] - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that [tex]\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}[/tex], [tex]\rho = 0.0899\,\frac{kg}{m^{3}}[/tex], [tex]L_{c} = 141790\,\frac{kJ}{kg}[/tex] and [tex]\dot W_{out} = 80\,kW[/tex], then the efficiency of this fuel cell is:

(Eq. 1)

[tex]\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)[/tex]

[tex]\dot W_{in} = 99.143\,kW[/tex]

(Eq. 2)

[tex]\eta = \frac{80\,kW}{99.143\,kW}[/tex]

[tex]\eta = 0.807[/tex]

The efficiency of this fuel cell is 80.69 percent.

How do I answer all the questions on this page?

Answers

Answer:

Create a google docs copy everything and paste hope this helps! :)

Explanation:

Expert Review is done by end users.

Answers

Answer:nononononono

Explanation:

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