## KJ4UTS one year ago What is the value of an investment of $6,835.69 at 3.88% interest compounded daily for 6 years? • This Question is Closed 1. KJ4UTS 2. Michele_Laino I'm sorry, I'm not good with financial mathematics 3. jebonna 1st year: 0.0388 x 6835.69 =$265.22 (2 decimal places) 6,835.69 + 265.22 = $7100.91 2nd year: 0.0388 x 7100.91 =$275.52 (2 decimal places) 7100.91 + 275.52 = $7376.43 3rd year: 0.0388 x 7376.43 =$286.21 (2 decimal places) 7376.43 + 286.21 = $7662.64 4th year: 0.0388 x 7662.64 =$297.31 (2 decimal places) 7662.64 + 297.31 = $7959.95 5th year: 0.0388 x 7959.95 =$308.85 (2 decimal places) 7959.95 + 308.85 = $8268.80 6th year: 0.0388 x 8268.80 =$320.83 (2 decimal places) 8268.80 + 320.83 = \$8589.63 Does this make sense? It's 3.88% interest on the new amount each year. I know this is closed, but I just thought I'd answer anyway haha. :)

4. jebonna

Im not sure if its right, considering it says daily o.o If not, I'm sorry xD But you might be able to work it out

5. IrishBoy123

Assuming the 3.88% is an annual rate, this is a pretty silly question the *wrong* way to calculate a daily rate r from an annual rate R is to say $$r = {R \over 365}$$ and so the answer to this question would be $$6,835.69 (1+{3.88\% \over 365})^{6 \times 365} = 8,627.42$$ the *correct* way is to say the daily $$r$$ follows from the fact that $$(1+r)^{365} = 1+ R$$ so $$r=(1+R)^{{1\over 365}}-1$$ but this means that the final amount $$A$$ of initial investment $$A_o$$ follows from $$A = A_o(1+r)^N = A_o((1+R)^{{1\over 365}})^N$$ where $$N = 6 \times 365$$ which is just $$A = A_o(1+R)^6 \quad \quad [= 6,835.69(1+3.88 \%)^6 = 8,589.62 ]$$ ie the correct way, decompounding for a daily rate and then compounding back, is a waste of time and you may as well just look at it annually.