An account is opened with $7,595.96 with a rate of increase of 2% per year. After 1 year, the bank account contains $7,746.90. Assuming no deposits or withdrawals are made, which equation can be used to find $y$, the amount of money in the account after $x$ years? (Round money values to the nearest penny.)
A. $y=7,746.90(1.02)^x$
B. $y=7,746.90(0.02)^x$
C. $y=7,595.96(1.02)^x$
D. $y=7,595.96(0.02)^x$
Answer (1)
The problem describes an exponential growth scenario.
The general form of the equation is y = P ( 1 + r ) x , where P is the initial amount and r is the rate of increase.
Substitute the given values: P = 7595.96 and r = 0.02 .
The equation is y = 7595.96 ( 1.02 ) x , so the answer is y = 7 , 595.96 ( 1.02 ) x .
Explanation
Understanding the Problem We are given an initial account balance of $7,595.96 with a rate of increase of 2% per year. We want to find an equation that models the amount of money, y , in the account after x years, assuming no deposits or withdrawals are made. The general form of an exponential growth equation is y = P ( 1 + r ) x , where P is the principal amount, r is the rate of increase, and x is the number of years.
Applying the Formula In this case, the principal amount P = 7595.96 and the rate of increase r = 0.02 . Substituting these values into the general equation, we get y = 7595.96 ( 1 + 0.02 ) x , which simplifies to y = 7595.96 ( 1.02 ) x .
Final Equation Therefore, the equation that can be used to find y , the amount of money in the account after x years, is y = 7 , 595.96 ( 1.02 ) x .
Examples
Exponential growth is a powerful tool in finance. For example, if you invest $1000 in a stock that grows at an average rate of 10% per year, you can use the equation y = 1000 ( 1.10 ) x to predict how much your investment will be worth after x years. This helps in making informed decisions about long-term investments and financial planning.
