A person invested $7,400 in an account growing at a rate allowing the money to
double every 13 years. How much money would be in the account after 11 years, to the
nearest dollar?

Respuesta :

Answer:

$13,304

Step-by-step explanation:

-We use the compound interest function to determine the rate of growth.

-Given that the amount doubles in 13 years, the annual growth  rate is calculated as:

[tex]A=P(1+i)^n\\\\2P=P(1+i)^n\\\\\therefore 2=(1+i)^{13}\\\\i=2^{1/13}-1=0.05477[/tex]

We now substitute this value of i in the compound interest formula equation to solve for future value:

[tex]A=P(1+i)^n\\\\\\=7400(1.05477)^{11}\\\\\\=13303.54[/tex]

[tex]\approx13304[/tex]

Hence, the future value to the nearest dollar is $13,304

*You can alternatively use the exponential growth function:

[tex]P_t=P_oe^{rt}\\\\2=e^{13r}\\\\r=\frac{In \ 2}{13}=0.05332\\\\\therefore P_{11}=7400e^{0.05332\times 11}\\\\=13303.14[/tex]

This is slightly off by just $1

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