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3 votes
A person invested $6,000 in an account growing at a rate allowing the money to

double every 11 years. How much money would be in the account after 19 years, to the
nearest dollar?

User Hunterjrj
by
7.3k points

1 Answer

2 votes

Answer:

P(19)=$19,852

To the nearest dollar.

Explanation:

Exponential Growth

The natural growth of some magnitudes can be modeled by the equation:


P(t)=P_o(1+r)^t

Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.

We are given the condition that an investment of Po=$6,000 in an account doubles every 11 years. The final value of the investment in t=11 is P(11)=$12,000.

Substituting into the general equation:


12,000=6,000(1+r)^(11)

Dividing by 6,000 and swapping sides:


(1+r)^(11)=2

Taking the 11th root:


1+r=\sqrt[11]{2}


1+r=1.065

Substituting into the formula:


P(t)=6,000(1.065)^t

Now we need to find the money in the account after t=19 years:


P(19)=$6,000(1.065)^(19)

P(19)=$19,852

To the nearest dollar.

User Sunding Wei
by
7.9k points

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