Someone Answer these Multiple choice questions

An investment of $57,800 in equipment has yearly profit of $25,000 for 6 years at an MARR of 6% and requires $12,300 yearly for its maintenance. Its salvage value is $17,600. The AW (MARR%) of this project is: (Hint: Use CR(6%) = $9,232.68 ) $2,198.5 O - $3,998.5 - $3,987.9 O $1,598.8 $3,467.32 O Other: The correct expression for given cash flow at N=O is: $182 $189 $196 $203 $210 $217 $224 $231 $238 $245 $252 $259 $175 1 2 3 4 5 6 7 00 9 10 11 12 13 $175(P/A,1%,13)+$7 (P/G,1%,13) $175+$182 (P/G,1%,12)+$7(P/A,1%,12) [$175(P/A,%,13)+$182 (P/A,1%,12)+$7(P/G,%,12)](P/F,1%,1) [$175+$182 (P/A,1%,12)+$7(P/G,1%,12)](P/F,1%,1) [$175+$182+$7(P/G,1%,12)](P/F,%,1) O Other: The value of P for given cash flow is: P $250 $150 i,=8.5% iz=10.5% iz=8.5% ix=6.5 % s=6.5% s=8.5% 0 1 2 3 4 5 6 CO $563 O $ 390 $ 264 $ 434 $454 $ 324 Other: Solve for F: 1 F $150 $250 $150 $250 in =8.5% i2=9.5% iz=8.5% 1456.5% s=7.5% 1s=5.5% 0 1 2 3 4 5 6 The expression for Effective Interest Rate (EIR) when the compounding period is daily and the interest rate is 10.5% 1.105^ 365 2.105-12 (1+0.105/365)^365 O ((1+0.105/12)^12) -1 1.105/365 O ((1+0.105/365)^365) - 1 Other: The economic notation for Gradient Series Compound Amount Factor is: O G (G/F, i%, N) O (F/G, i%, N) O (G/P, 1%, N) (P/F, 1%, N) (P/A, i%, N) O Other: If C= $15,000, N= 20, Z=$15,000, i= 20%, r=8%. If the payments are made semiannually then VN will be: (Hint: Use (P/F, 10%, 40) = 0.221 and (P/A, 10%, 40) = 9.779) O $8765 $4573 Rs. 567345 $9182 $5673 Other