Question

1 year= $3000

X Problem 6-2 (algorithmic) Question Help O $10,000 OOO Years 0 ! 53.00 53.000 $.000 $2,000 $12.000 315.000 Click the icon to view the interest factors for discrete compounding when = 7% per year. i More Info The equivalent annual worth is S . (Round to the nearest dollar) Single Payment Compound Present Amount Worth Factor Factor (F/P.I, NJ (P/F, I, NJ 1.0700 0.9346 1.1449 0.8734 1.2250 0.8163 1.3108 0.7629 1.4028 0.7130 Compound Amount Factor (F/A, i, N) 1.0000 2.0700 3.2149 4.4399 5.7507 Equal Payment Series Sinking Present Fund Worth Factor Factor (A/F, I, N) (P/A, N) 1.0000 0.9346 0.4831 1.8080 0.3111 2.6243 0.2252 3.3872 0.1739 4.1002 Capital Recovery Factor (A/P, IN) 1.0700 0.5531 0.3811 0.2952 0.2439 Gradient Series Gradient Gradient Uniform Present Series Worth (A/G, i, N) (P/G, I, N) 0.0000 0.0000 0.4831 0.8734 0.9549 2.5060 1.4155 4.7947 1.8650 7.6467 OWN 1.5007 1.6058 1.7182 1.8385 1.9672 0.6663 0.6227 0.5820 0.5439 0.5083 7.1533 8.6540 10.2598 11.9780 13.8164 0.1398 0.1156 0.0975 0.0835 0.0724 4.7665 5.3893 5.9713 6.5152 7.0236 0.2098 0.1856 0.1675 0.1535 0.1424 2.3032 2.7304 3.1465 3.5517 3.9461 10.9784 14.7149 18.7889 23.1404 27.7156 Print Done Enter your answer in the answer box and then click Check Answer. All parts showing Clear All Check Answer

2 year= $3000

3 year= $6000

4 year= $9000

5 year= $12000

6 year= $15000

Answer 1

Ans: -$3721

Explanation:

Here, P = 18,000

          A1 = 3,000

          G = 3,000

           i = 7%

           n = 6

Present Worth (PW) = 18,000 - 3000(P/F, 7%, 1) - 3000(P/F, 7%, 2) - 6000(P/F, 7%, 3) - 9000(P/F, 7%, 4) - 12,000(P/F, 7%, 5) - 15000(P/F, 7%, 6).

         = 18,000 - 3000(0.9346) - 3000(0.8734) - 6000(0.8163) - 9000(0.7629) - 12,000(0.7130) - 15000(0.6663)

         = 18,000 - 2803.8 - 2620.2 - 4897.8 - 6866.1 - 8556 - 9994.5

         = -17,738.4

EAW = PW(A/P, i, n)

         = -17,738.4(A/P, 7%, 6)

         = -17,738.4(0.2098)

         = -$3,721