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You place $100 per month into an account that earns 1% per month. Which of the following expressions can be used to calculate the account's value after 3 years? P = 100(P/A, 1%, 3) F = 100 (P/A, 1%, 36) F = 100[(1 + 0.01)n - 1]/0.01 F = 100(F/A, 12.68%, 3) Today $5000 is deposited in an account that cams 2.5% per quarter. Additional deposits are made at the end of every quarter for the next 20 years. The deposits start at $100 and increase by S50 each year thereafter. The amount that has accumulated in this account (FW) at the end of 20 years can be expressed as follows. FW = 5000(P/F, 2.5%, 20) + 100(F/A, 2.5%, 20%) + 50 (P/G, 2.5%, 20-1) FW = 5000(F/P, 10%, 80) + 150(P/G, 10%, 80) (F/P, 10%, 80) FW = 5000(P/F, 10.38%, 20) + 100(P/A, 10.38%, 20) + 50(P/G, 10.38%, 20) (P/F, 10.30%, 1) FW = 5000(F/P, 10.38%, 80) + 100(F/A, 2.5%, 80) +50(P/G, 2.5%, 80)(F/P, 2.5%, 80) A project is being considered that has a First cost of $12, 500, creates $5000 in annual cost savings, requires $3000 in annual operating costs, and has a salvage value of $2000 after a project life of 3 years. If interest is 10% per year, which formula calculates the project's present worth? PW = 12, 500(P/F, 10%, 1) + (-5000 + 3000)(P/A, 10%, 3) - 2000 (F/P, 10%, 3) PW = -12, 500 + (5000 - 3000) (P/A, 10%, 3) - 2000(P/F, 10%, 3) PW= 12, 500(F/P, 10%, 3) + (5000 - 3000)(F/A, 10%, 3) + 2000 PW = -12, 500 + 5000(P/A, 10%, 3) - 3000(P/A, 10%, 3) + 2000(P/F, 10%, 3)