Economy with Social and Global Implications

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cbv1234

Economics

usf

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I have about 7 questions, and I need you to help me to answer them. The course name is Engineering Economy with Social and Global Implications (EGN3615). I will upload the questions and sample of the answers. PLease, i want you to answer them in Excel.

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The attached excel file, Econ HW.xlsx , is where you are to do your Economics homework. It contains your instructions. The exercises, sometimes called problems, that you are to solve are listed here. * Problems pp. 99 ff: 3-2, 3-3, 3-20, 3-21, 3-26, 3-27, 3-32 Do your work in the Excel file and submit it to this assignment. Be sure to state which week’s assignment it is and name each exercise. Chapter 3 3-5 a) $1,500,000 = P(1+0.1*2) = 1.2P P = $1,500,000/1.2 = $1,250,000 b) $1,500,000 = P(1+0.1)2 = $1,500,000 (1.210) P = $1,500,000/1.210 = $1,239,669 3-6 a) P = P + P*i*n 2 = 1 + 0.03*n n = 1/0.03 = 33.3 years b) 2P = P(1+ 0.03)n 2 = 1.03n ln(2) = n*ln(1.03) n = ln(2)/ln(1.03) = 23.45 years 3-10 a) Interest Rates i. Interest rate for the past year = ($100 – $90)/$90 = $10/$90 = 0.111 or 11.1% ii. Interest rate for the next year = ($110 – $100)/$1 = 0.10 or 10% b) $90 (F/P, i%, 2) = $110 (F/P, i%, 2) = (1 + i)2= $110/$90= 1.222 (1+i) = sqrt(1.222) = 1.1054 i= 1.1054 – 1 = 0.1054 = 10.54% 3-20 Use 6000 = 5000 (F/P, i, n) = 5000 (1 + i)n. a) n = 2, 𝑖𝑖 = √1.2 − 1 = 0.0954 or 9.54% 3 b) n = 3, 𝑖𝑖 = √1.2 − 1 = 0.0627 or 6.27% 5 c) n = 5, 𝑖𝑖 = √1.2 − 1 = 0.0371 or 3.71% 10 d) n = 10, 𝑖𝑖 = √1.2 − 1 = 0.0184 or 1.84% 3-29 Calculator Solution 1% per month F = $1,000 (1 + 0.01)12 = $1,126.83 12% per year F = $1,000 (1 + 0.12)1 = $1,120.00 Savings in interest = $6.83 Compound interest table solution 1% per month F = $1,000 (1.127) = $1,127.00 12% per year F = $1,000 (1.120) = $1,120.00 Savings in interest = $7.00 3-37 F16 = $10,000 (1 + 0.025/4)16 = $11,048.27 F40 = $11,048.27 (1 + 0.035/4)24 = $13,617.56 3-40 Effective Interest Rate = (1 + i)m − 1 = (1 + 0.09/4)4 − 1 = 0.0931= 9.31% 3-44 Effective interest rate = (1 + r/m)m − 1 .161 = (1 + r/12)12 - 1 (1 + r/12) = 1.1610.0833 = 1.0125 r/12 = .0125 r = 12(0.0125) = 15.0% 3-59 a) 11.98% compounded continuously F = $10,000 e(0.1198)(4) = $16,147.82 b) 12% compounded daily F = $10,000 (1 + 0.12/365)365*4 = $16,159.47 c) 12.01% compounded monthly F = $10,000 (1 + 0.1201/12)12*4 = $16,128.65 d) 12.02% compounded quarterly F = $10,000 (1 + 0.1202/4)4*4 = $16,059.53 e) 12.03% compounded yearly F = $10,000 (1 + 0.1203)4 = $15,752.06 Decision: Choose Alternative (b)
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Explanation & Answer

Attached.

Engineering Economy
3_2

Interest = principal *Rate* Time
$ 2000 * 0.1 * 3 years
$600
Amount= principal +interest
$2000+600
$2,600

3_3

Interest = Principal *Rate*Time
$350 = $5000 * 0.08 * Time
$350 = 400T
Time =350/4...

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