Question

precal application problem ***please help me!***

One freezer costs $623.95 and uses a 90 kilowatt hours (kwh)of electricity each month. A second freezer costs $500 and uses 100kwh of electricity each month. Theexpected life of each freezer is12 years. What is the minimum electric rate in cents per kwh forwhich the 12-year total cost (purchase price + electricitycosts)will be less for the first freezer?***Has something to do with inequalities, but I'm just reallystuck.
0 0
Add a comment Improve this question Transcribed image text
Answer #1

First you need to assume the freezers are never unplugged, otherwise you will get a different answer.

Over 12 years, you have 144 months.

Assume electric rate = Y

Cost of first freezer = initial cost plus monthly cost = $723.95 + 90 kwh * 144 * Y

Cost of second freezer = initial cost plus monthly cost = $600 + 100 kwh * 144 * Y

These two costs are equal when 723.95 + 12960Y = 600 + 14400Y or 123.95 = 1440Y or Y = 8.61 cents, therefore this is the minimum cost where freezer 1 is less.

Add a comment
Know the answer?
Add Answer to:
precal application problem ***please help me!***
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Problem 5-49 (algorithmic) Show Work | Question Help In this problem, we consider replacing an ex...

    Problem 5-49 (algorithmic) Show Work | Question Help In this problem, we consider replacing an existing electrical water heater with an array of solar panels. The net installed investment cost of the panels is $1,520 ($1,900 less a 20% tax credit from the government). Based on an energy audit, the existing water heater uses 200 kilowatt hours (kWh) of electricity per month, so at $0.14 per kWh, the cost of operating the water heater is $28 per month. Assuming the...

  • Problem 6-16 (algorithmic) Question Help A special-purpose 30-horsepower electric motor has an efficiency of 90%. Its...

    Problem 6-16 (algorithmic) Question Help A special-purpose 30-horsepower electric motor has an efficiency of 90%. Its purchase and installation price is $2,100. A second 30-horsepower high-efficiency motor can be purchased for $3,100, and its efficiency is 94%. Either motor will be operated 3,500 hours per year at full load, and electricity costs $0.09 per kilowatt-hour (kWh). MARR = 12% per year, and neither motor will have a market value at the end of the eight-year study period a. Which motor...

  • In this​ problem, we consider replacing an existing electrical water heater with an array of solar...

    In this​ problem, we consider replacing an existing electrical water heater with an array of solar panels. The net installed investment cost of the panels is ​$1,365 ​($2,100 less a 35​% tax credit from the​ government). Based on an energy​ audit, the existing water heater uses 180 kilowatt hours​ (kWh) of electricity per​ month, so at ​$0.13 per​ kWh, the cost of operating the water heater is ​$23.4 per month. Assuming the solar panels can save the entire cost of...

  • In this problem, we consider replacing an existing electrical water heater with an array of solar...

    In this problem, we consider replacing an existing electrical water heater with an array of solar panels... Answer all parts of question if there is multiple parts not just what's below the line please ;) In this problem, we consider replacing an existing electrical water heater with an array of solar panels. The net installed investment cost of the panels is $1,080 ($1,800 less a 40% tax credit from the government). Based on an energy audit, the existing water heater...

  • Need help with this please Lab 14 Electricity Bill Worksheet Background knowledge. The power of an...

    Need help with this please Lab 14 Electricity Bill Worksheet Background knowledge. The power of an electrical appliance is measured in watt (W). One watt is a measure of the energy used up in one second. The unit used for calculating power consumption is the kilowatt-hour (kWh). In order to calculate power consumed by an electrical appliance, we first convert the power to kilowatt (kW). For example, 800 W=0.8 kW, 2000 W = 2.0 kW and so on. One unit...

  • In this​ problem, we consider replacing an existing electrical water heater with an array of solar...

    In this​ problem, we consider replacing an existing electrical water heater with an array of solar panels. The net installed investment cost of the panels is ​$1,235 ​($1,900 less a 35​% tax credit from the​ government). Based on an energy​ audit, the existing water heater uses 210 kilowatt hours​ (kWh) of electricity per​ month, so at ​$0.14 per​ kWh, the cost of operating the water heater is ​$29.4 per month. Assuming the solar panels can save the entire cost of...

  • Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to...

    Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to generate electricity and to reduce the electricity costs for their offices, which are located in Toowoomba. Currently the business uses 60,000 kilowatt hours (kWh) per quarter (3 months) at an average cost of $0.30 per kwh, supplied by the local coal fired power station. The current required rate of return used to evaluate projects is 6%, with a required payback period of 3 years....

  • Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to...

    Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to generate electricity and to reduce the electricity costs for their offices, which are located in Toowoomba. Currently the business uses 60,000 kilowatt hours (kWh) per quarter (3 months) at an average cost of $0.30 per kwh, supplied by the local coal fired power station. The current required rate of return used to evaluate projects is 6%, with a required payback period of 3 years....

  • Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to...

    Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to generate electricity and to reduce the electricity costs for their offices, which are located in Toowoomba. Currently the business uses 60,000 kilowatt hours (kWh) per quarter (3 months) at an average cost of $0.30 per kwh, supplied by the local coal fired power station. The current required rate of return used to evaluate projects is 6%, with a required payback period of 3 years....

  • Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to...

    Zephyr Farming Pty Ltd is considering the purchase of a wind turbine generator in order to generate electricity and to reduce the electricity costs for their offices, which are located in Toowoomba. Currently the business uses 60,000 kilowatt hours (kWh) per quarter (3 months) at an average cost of $0.30 per kwh, supplied by the local coal fired power station. The current required rate of return used to evaluate projects is 6%, with a required payback period of 3 years....

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT