Consider the cash flows shown below. Assuming an annual interest rate of 10%, What is the present worth for the best alternative Data Equipment Equipment Y $. Initial Cost $40,000 $70,000 Annual Maintenance Cost, $ per year $5,000 $1,000 $ Salvage value $4,000 $2,000 Life, Years 4 8 APPENDIX C: COMPOUND INTEREST TABLES 613 1096 10% Sinking Fund Factor Find a Given F AFF 100 14752 Single Payment Compound Present AMOUISE Worth Factor Factor Find Find P Given P Given FP PE NICO 19091 112 S2 13 7513 1464 6830 1.6TI 1620 171 5805 1940 SI 2144 4 2.358 241 2 n Present Worth Factor Find P Given A PA 0.900 1.736 20187 0170 1 2 Compound Interest Factors Uniform Payment Series Capital Compound Recovery Amount Factor Factor Find A Find Given P Given A FA 1.10 11000 2.100 4021 13210 24 12638 OS 7718 23 9.157 17 36 19736 1979 20 03 5762 Arithmetic Gradient Gradient Gradient Uniform Present Series Worth Find A Find P Given G Given G A/G P/G 0 0 076 0.826 0.037 2329 1381 4.378 1.810 6.862 2.294 9.681 . 12.763 3 16:20 37 19.121 Us 20 891 1 2 3 - 3 4 1 5 1791 5 2133 168 0206 TOS 091 07016 7 8 2 TO 4355 180 5335 1750 60145 6 7 8 10 -S2 The -5654 S159 445. -574 nas