114k views
21 votes
EMERGENCY PLS HELP: A person invested $420 in an account growing at a rate allowing the money to double every 7 years. How much money would be in the account after 16 years, to the nearest dollar?

User Zrg
by
8.0k points

2 Answers

9 votes

Final answer:

To calculate the growth of a $420 investment that doubles every 7 years after 16 years, find the full doubling periods and the growth for the remaining time, which leads to approximately $2094 in the account after 16 years.

Step-by-step explanation:

Calculation of Investment Growth

The amount of money in an account after a certain number of years can be calculated using the formula for compound interest. As the money doubles every 7 years, this represents an exponential growth according to the rule of 72. To find the total after 16 years, we divide 16 by 7 to calculate the number of full doubling periods, which in this case is about 2.29 periods. Since the investment is not exactly at a doubling point at the 16-year mark, we can use the power of compounding to estimate the final amount.


For the first 14 years (or two doubling periods), the $420 investment will double twice:

After 7 years: $420 × 2 = $840

After 14 years: $840 × 2 = $1680

For the remaining 2 years, we need to find the growth factor for a 7-year period (which is 2) and then raise it to the fraction of time (2/7 years) that we're interested in. This can be represented as:

$1680 × 2^(2/7) = $1680 × approximately 1.246 = $2094

Rounding this to the nearest dollar, we have $$2094 in the account after 16 years.

User Mussnoon
by
8.6k points
6 votes

Answer:

2048

Step-by-step explanation:

a=420

d=7-- d is the doubling time

t=16

plug in y =420(2) 16/7

y=2047.94293906

y≈2048

User KotoroShinoto
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories