Answer:
Part a) [tex]\$3,826.79[/tex]
Part b) [tex]\$3,858.74[/tex]
Step-by-step explanation:
The complete question is
In 1884, a person sold a house to a lady for $26. If the lady had put the $26 into the bank account paying 4% interest, how much would the investment have been worth in the year 2009 if interest were compounded in the following ways?
a) monthly b) continuously
Part a) compounded monthly
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=2009-1884=125\ years\\ P=\$26\\ r=4\%=4/100=0.04\\n=12[/tex]
substitute in the formula above
[tex]A=26(1+\frac{0.04}{12})^{12*125}[/tex]
[tex]A=26(\frac{12.04}{12})^{1,500}[/tex]
[tex]A=\$3,826.79[/tex]
Part b) compounded continuously
we know that
The formula to calculate continuously compounded interest is equal to
[tex]A=P(e)^{rt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
e is the mathematical constant number
we have
[tex]t=2009-1884=125\ years\\ P=\$26\\ r=4\%=4/100=0.04[/tex]
substitute in the formula above
[tex]A=26(e)^{0.04*125}[/tex]
[tex]A=26(e)^{5}[/tex]
[tex]A=\$3,858.74[/tex]
