A person invested $9,000 in an account growing at a rate allowing the money to double every 10 years. How much money would be in the account after 19 years, to the nearest dollar? Answer:

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A person invested  $9,000 in an account growing at a rate allowing the money to double every 10 years. How much money would be in the account after 19 years, to the nearest dollar? Answer:

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Initial Investment and Doubling Period: Determine the initial investment and the doubling period. The initial investment (a) is $9,000, and the money doubles every 10 years (T).Number of Doubling Periods: Determine the number of times the initial investment will double in 19 years. The number of doubling periods (x) is calculated by dividing the total time (t) by the doubling period (T). x = t / T = 19 / 10 = 1.9 Since the investment can only double a whole number of times, we need to use the floor function to find the number of complete doubling periods, which is 1.Amount After Complete Doubling Periods: Calculate the amount of money after the complete doubling periods. The formula for exponential growth is P(x) = a * (b^x), where P(x) is the final amount, a is the initial amount, b is the growth factor (2 for doubling), and x is the number of doubling periods. P(1) = 9000 * (2^1) = 9000 * 2 = 18000 After one complete doubling period (10 years), the investment will be $18,000.Growth for Remaining Years: Calculate the growth for the remaining years. There are 9 years left after the first complete doubling period. We need to calculate the growth for these 9 years. The growth rate per year (r) can be calculated using the formula for doubling time: T = ln(2) / ln(1 + r), where ln is the natural logarithm. We need to solve for r, but since we already know the investment doubles every 10 years, we can use the rule of 72, which is an approximation that states T ≈ 72 / r. Therefore, r ≈ 72 / T = 72 / 10 = 7.2% per year.Apply Annual Growth Rate: Apply the annual growth rate for the remaining 9 years. The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, including interest, P is the principal amount, r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the time the money is invested for. Since the interest is compounded once per year in this case (n = 1), the formula simplifies to A = P(1 + r)^t. A(9) = 18000 * (1 + 0.072)^9Final Amount After 19 Years: Calculate the final amount after 19 years. A(9) = 18000 * (1 + 0.072)^9 A(9) ≈ 18000 * (1.072)^9 A(9) ≈ 18000 * 1.872 A(9) ≈ 33729.6 Rounding to the nearest dollar, the final amount is approximately $33,730.

A person invested $\$ 9,000$ in an account growing at a rate allowing the money to double every $10$ years. How much money would be in the account after $19$ years, to the nearest dollar? $\newline$ Answer:

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A person invested $\$ 9,000$ in an account growing at a rate allowing the money to double every $10$ years. How much money would be in the account after $19$ years, to the nearest dollar? $\newline$ Answer: