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A person invested
$1,100 in an account growing at a rate allowing the money to double every 15 years. How much money would be in the account after 6 years, to the nearest dollar?
Answer:

A person invested $\$ 1,100$ in an account growing at a rate allowing the money to double every $15$ years. How much money would be in the account after $6$ years, to the nearest dollar? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A person invested

\$ 1,100

in an account growing at a rate allowing the money to double every

15

years. How much money would be in the account after

6

years, to the nearest dollar?

\newline

Answer:

Identify Growth Type: Determine the type of growth. $\newline$ The account grows at a rate that allows the money to double every $15$ years. This indicates exponential growth.
Find Initial Amount and Factors: Identify the initial amount $a$ , the growth factor $b$ , and the time period for doubling $T$ . The initial amount $a$ is $\$1,100$ , the growth factor $b$ is $2$ (since the money doubles), and the time period for doubling $T$ is $15$ years.
Calculate Doubling in $6$ Years: Calculate the number of times the initial amount will double in $6$ years. $\newline$ We need to find $x$ , the fraction of the $15$ -year period that has passed in $6$ years. $\newline$ $x = \frac{t}{T}$ where $t$ is $6$ years and $T$ is $15$ years. $\newline$ $x = \frac{6}{15} = 0.4$
Use Exponential Growth Formula: Use the exponential growth formula to calculate the final amount. $\newline$ The exponential growth formula is $P(x) = a(b)^{(x)}$ . $\newline$ Substitute $\$1,100$ for $a$ , $2$ for $b$ , and $0.4$ for $x$ . $\newline$ $P(0.4) = 1100(2)^{(0.4)}$
Evaluate Final Amount: Evaluate the expression to find the final amount. $\newline$ $P(0.4) = 1100(2)^{0.4}$ $\newline$ To calculate $2$ raised to the power of $0.4$ , we can use a calculator. $\newline$ $2^{0.4} \approx 1.3195$ $\newline$ Now multiply this by $\$1,100$ . $\newline$ $P(0.4) \approx 1100 \times 1.3195 \approx 1451.45$ $\newline$ Round to the nearest dollar. $\newline$ $P(0.4) \approx \$1,451$

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