The following data is available for three different alternatives. Assume an interest rate of 10% per year, compounded annually. Initial Cost Annual Benefit Useful Life (yrs) Alternative A 10,000 1,450 infinite Alternative B 11,000 1,930 10 Alternative C 20,000 6,707 5 Alternatives B and C are replaced at the end of their useful lives with identical replacements. Using present worth analysis, find the best alternative.

If you could write out the equation once for me to see where the values should go, that would be helpful. Thank you!

Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $20,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle's MARR is 10%/year. Part a What is the future worth of this investment? $ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is £10.

Given the data for three different alternatives in the table below, determine the best alternative using the incremental rate of return (∆RoR) analysis. MARR =9%.     A B C First cost $15,000 $25,000 $20,000 O &M Cost/ year 1,600 400 900 Benefit/year 8,000 13,000 9,000 Salvage value 3,000 6,000 4,600 Life in years 4 4 4   The better alternative between the second  increment (B – C) is:. Group of answer choices Alt. B Alt. C Alt. B or Alt. C Alt. A

Most likely estimates for a project are as follows. MARR Useful life Initial investment Receipts - Expenses (R-E) 10% per year 9 years $5,000 $1,200/year Click the icon to view the relationship between the PW and the percent change in parameter. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. (b) To which variable is the PW most sensitive to? OA. Receipts - Expenses OB. Usefule life OC. Initial Investment (a) Determine whether the statement "An initial investment of $6,000 keeps the investment economical." is true or false. O False O True