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

an initial investment amount p, an annual interest rate r, and a time t are given. find the future value of the investment when interest is compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. then find (e) the doubling - time t for the given interest rate. p = $350, r = 2.00%, t = 13 yr. a) the future value of the investment when interest is compounded annually is $ (type an integer or a decimal. round to the nearest cent as needed.) b) the future value of the investment when interest is compounded monthly is $ (type an integer or a decimal. round to the nearest cent as needed.) c) the future value of the investment when interest is compounded daily is $ (type an integer or a decimal. round to the nearest cent as needed.) d) the future value of the investment when interest is compounded continuously is $ (type an integer or a decimal. round to the nearest cent as needed.) e) find the doubling - time for the given interest rate. t = yr (type an integer or decimal rounded to two decimal places as needed.)
Answer
Explanation:
Step1: Identify the compound - interest formulas
The compound - interest formula for compounded $n$ times a year is $A = P(1+\frac{r}{n})^{nt}$, and for continuous compounding is $A = Pe^{rt}$. Here, $P=$350$, $r = 0.02$ (since $2.00%=0.02$), and $t = 13$ years.
Step2: Calculate future value when compounded annually ($n = 1$)
$A=P(1 + r)^{t}$. Substitute $P = 350$, $r=0.02$, and $t = 13$ into the formula: $A=350\times(1 + 0.02)^{13}=350\times1.02^{13}$. Using a calculator, $1.02^{13}\approx1.293607$, so $A = 350\times1.293607\approx452.76$.
Step3: Calculate future value when compounded monthly ($n = 12$)
$A=P(1+\frac{r}{n})^{nt}$. Substitute $P = 350$, $r = 0.02$, $n = 12$, and $t = 13$ into the formula: $A=350\times(1+\frac{0.02}{12})^{12\times13}=350\times(1+\frac{0.02}{12})^{156}$. $1+\frac{0.02}{12}\approx1.001667$, $(1.001667)^{156}\approx1.299977$, so $A = 350\times1.299977\approx454.99$.
Step4: Calculate future value when compounded daily ($n = 365$)
$A=P(1+\frac{r}{n})^{nt}$. Substitute $P = 350$, $r = 0.02$, $n = 365$, and $t = 13$ into the formula: $A=350\times(1+\frac{0.02}{365})^{365\times13}=350\times(1+\frac{0.02}{365})^{4745}$. $1+\frac{0.02}{365}\approx1.0000548$, $(1.0000548)^{4745}\approx1.300947$, so $A = 350\times1.300947\approx455.33$.
Step5: Calculate future value when compounded continuously
$A = Pe^{rt}$. Substitute $P = 350$, $r = 0.02$, and $t = 13$ into the formula: $A=350\times e^{0.02\times13}=350\times e^{0.26}$. Since $e^{0.26}\approx1.309964$, $A = 350\times1.309964\approx458.49$.
Step6: Find the doubling - time
For continuous compounding, we use the formula $A = Pe^{rt}$. If $A = 2P$, then $2P=Pe^{rt}$. Canceling out $P$ (since $P\neq0$), we get $2=e^{rt}$. Taking the natural logarithm of both sides, $\ln(2)=\ln(e^{rt})=rt$. Solve for $t$: $t=\frac{\ln(2)}{r}$. Substitute $r = 0.02$ into the formula, $t=\frac{\ln(2)}{0.02}=\frac{0.693147}{0.02}\approx34.66$.
Answer:
a) $452.76$ b) $454.99$ c) $455.33$ d) $458.49$ e) $34.66$