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題目中的 $ 可能是指 港幣, 請參照意見欄. 計算再修正如下. Sol : plan a : r = 8% / 2 = 0.08 / 2 = 0.04 半年記息一次, 故 2 年記息 4 期. A( plan a ) = 20000 * ( 1 + 0.04 )^4 plan b : r = 10% = 0.1 每年記息一次, 故 2 年記息 2 期. 設 2 年後的匯率為 1英鎊 兌換 k 元. 因為第 0 年換成英鎊(12...
Sol 10000*(1+0.04)^3 =10000*1.1248 =11248
1a) 5000*(1+0.095)^17 =$23888.9 ∴It will grow to $23888.9 at the end of 17 years. 1b) Let n be the year that the money grow to $18000. 5000*(1+0.095)^n = 18000 (1.095)^n = 3.6 log (1.095)^n = log 3.6 n log (1.095) = log 3.6 n=14.1 ∴It takes 15 years for the money grow to $18000...
1. 1) S : client investing in the stock market B : client investing in the bond market P...
Amount = 70000 x (1 + 6%/12)^8 Interest = 70000 x (1 + 6%/12)^8 - 70000 Amount = 8000 x (1 + 10%/12)^3 Interest = 8000 x (1 + 10%/12)^3 - 8000
This is a traditional setting in statistics of investment risk. The model is about the Markowitz's Modern Portfolio Theory. 圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20140220135841.jpg
The total value after 4 years= 20000 * (1 + 4%)^16= 37459.6249Let each cash flow is xx + x(1 + 4%) + ... + x(1 + 4%)^15 = 37459.6249x[(1.04)^16 - 1]/0.04 = 37459.624921.8245x = 37459.6249x = 1716.4
After first year, the fund balance is 10000*(1+r), not the one you said. For t=2, balance = 10000(1.05)(1+r) + 10000*(1+r)^2 For t=3, balance = 10000(1.05)^2(1+r) + 10000(1.05)(1+r)^2 + 10000(1+r)^3 ... For t=n, balance = 10000[(1.05)^(n-1)](1+r) + 10000[(1.05)^(n-2)](1+r)^2 + 10000[(1.05)^(n-3...
...: The $20000 deposited at time 0 has been invested for 5 years, so it would accumulate to $20000(1.06)⁵ at the end of year...
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