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Sagot :
To solve the question, we would be making use of the compound interest formula. This is given as;
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]A=final amount = $20,000
P=initial principal balance
r=interest rate =3%
n=number of times interest applied per time period
t=number of time periods elapsed = 6 years
Part A
If the interest is compounded annually, n =1
Therefore;
[tex]\begin{gathered} 20000=P(1+\frac{0.03}{1})^{1\times6} \\ 20000=P(1+0.03)^6 \\ (1.03)^6P=20000 \\ P=\frac{20000}{(1.03)^6} \\ P=16749.69 \end{gathered}[/tex]Answer: The minimum amount would be $16749.69
Part B
If the interest is compounded monthly, n =12
[tex]\begin{gathered} 20000=P(1+\frac{0.03}{12})^{6\times12} \\ 20000=P(\frac{12+0.03}{12})^{72} \\ (\frac{12.03}{12})^{72}P=20000 \\ P=\frac{20000}{(\frac{12.03}{12})^{72}} \\ P=16709.16 \end{gathered}[/tex]
Answer: The minimum amount would be $16709.16
Part C
If the interest is compounded monthly, n =365
[tex]\begin{gathered} 20000=P(1+\frac{0.03}{365})^{365\times6} \\ 20000=P(\frac{365+0.03}{365})^{2190} \\ (\frac{365.03}{365})^{2190}P=20000 \\ P=\frac{20000}{(\frac{365.03}{365})^{2190}} \\ P=16705.53 \end{gathered}[/tex]Answer: The minimum amount would be $16705.53
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