Engineering Economics What is normally used to compare alternatives that accomplish the same purpose but have unequal lives? Annual cost method Capitalized cost method MARR Present worth method Annual cost method Capitalized cost method MARR Present worth method ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics The alternatives which are standalone solutions for given situations in engineering involve: All of these A purchase cost (first cost) The anticipated resalable value (salvage value) and the interest return (rate of return) The anticipated life of the assets All of these A purchase cost (first cost) The anticipated resalable value (salvage value) and the interest return (rate of return) The anticipated life of the assets ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics What refers to the residual value of a company’s assets after all outside liabilities (shareholders excluded) have been allowed for? Return Equity Dividend Par value Return Equity Dividend Par value ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A factory operator bought a diesel generator set for P 10,000.00 and agreed to pay the dealer uniform sum at the end of each year for 5 years at 8% interest compounded annually, that the final payment will cancel the debt for principal and interest. What is the annual payment? P 2,540.56 P 2,504.57 P 2,544.45 P 2,500.57 P 2,540.56 P 2,504.57 P 2,544.45 P 2,500.57 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics The exact simple interest of P5,000 invested from June 21, 1995 to December 25, 1995 is P100. What is the rate of interest? 0.0398 0.0395 0.0392 0.039 0.0398 0.0395 0.0392 0.039 ANSWER DOWNLOAD EXAMIANS APP
Engineering Economics A man invested P110,000 for 31 days. The net interest after deducting 20% withholding tax is P890.36. Find the rate of return annually. 0.115 0.1195 0.1232 0.1175 0.115 0.1195 0.1232 0.1175 ANSWER DOWNLOAD EXAMIANS APP
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