an initial investment amount $p$, an annual interest rate $r$, and a time $t$ are given. find the future…

Explanation:

Step1: Recall compound - interest formulas

• Annual compounding: The formula for compound - interest is $A = P(1 + r)^t$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the number of years.
• Given $P=$350$, $r = 0.02$ (since $2.00%=0.02$), and $t = 13$ years. $A = 350(1 + 0.02)^{13}$ $A=350\times1.02^{13}$ Using a calculator, $1.02^{13}\approx1.293607$, so $A = 350\times1.293607\approx452.76$

Step2: Recall monthly compounding formula

The formula for compound - interest compounded $n$ times a year is $A=P(1+\frac{r}{n})^{nt}$. For monthly compounding, $n = 12$. $A = 350(1+\frac{0.02}{12})^{12\times13}$ $A = 350(1+\frac{0.02}{12})^{156}$ $1+\frac{0.02}{12}\approx1.001667$, and $(1.001667)^{156}\approx1.29997$. So $A = 350\times1.29997\approx454.99$

Step3: Recall daily compounding formula

For daily compounding, assume $n = 365$. $A=P(1 +\frac{r}{365})^{365t}$ $A = 350(1+\frac{0.02}{365})^{365\times13}$ $1+\frac{0.02}{365}\approx1.0000548$, and $(1.0000548)^{4745}\approx1.30094$. So $A = 350\times1.30094\approx455.33$

Step4: Recall continuous - compounding formula

The formula for continuous compounding is $A = Pe^{rt}$. $A=350\times e^{0.02\times13}$ $A = 350\times e^{0.26}$ Since $e^{0.26}\approx1.30996$, then $A = 350\times1.30996\approx458.49$

Step5: Find the doubling - time formula

For continuous compounding, if $A = 2P$, then $2P=Pe^{rt}$. Canceling out $P$ gives $2 = e^{rt}$. Solving for $t$ (the doubling - time $T$), we take the natural logarithm of both sides: $\ln(2)=\ln(e^{rt})$. Since $\ln(e^{rt})=rt$, then $T=\frac{\ln(2)}{r}$. $T=\frac{\ln(2)}{0.02}\approx\frac{0.693147}{0.02}=34.66$ years

Answer:

• (a) Annual compounding: $$452.76$
• (b) Monthly compounding: $$454.99$
• (c) Daily compounding: $$455.33$
• (d) Continuous compounding: $$458.49$
• Doubling - time: $34.66$ years