An account is opened with \[tex]$7,595.96 at an interest rate of 2% per year. After 1 year, the bank account contains \$[/tex]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. [tex]\( y = 7,595.96(0.02)^x \)[/tex]



Answer :

First, let's identify the initial amount in the account. The initial amount is [tex]$\$[/tex] 7,595.96$.

Next, we are given that the rate of increase per year is [tex]$2\%$[/tex]. This rate can be expressed as a decimal for calculations, which is [tex]$1.02$[/tex] (since [tex]$2\%$[/tex] increase means multiplying by [tex]$1 + \frac{2}{100} = 1.02$[/tex]).

We need to find the equation that accurately represents the amount of money in the account, [tex]$y$[/tex], after [tex]$x$[/tex] years of continuous growth without any deposits or withdrawals.

The general form for compound interest (where interest is compounded annually) is:
[tex]\[ y = P \times (1 + r)^x \][/tex]
where:
- \( P \) is the principal amount (initial amount),
- \( r \) is the annual growth rate as a decimal,
- \( x \) is the number of years.

Substituting the given values:
- The initial principal amount \( P \) is [tex]$\$[/tex] 7,595.96$,
- The annual growth rate \( r \) is [tex]$0.02$[/tex], which translates to multiplying by [tex]$1.02$[/tex].

Thus, the equation to find \( y \), the amount of money in the account after \( x \) years, is:
[tex]\[ y = 7,595.96 \times (1.02)^x \][/tex]

Therefore, the correct choice is:
[tex]\[ \boxed{y = 7,595.96(1.02)^x} \][/tex]

This matches the third equation listed in the options. Therefore, you should choose:
[tex]\[ y = 7,595.96(1.02)^x \][/tex]
In conclusion, the correct equation is option number 3.