an account is opened with $7,595.96 with a rate of increase of 2% per year. after 1 year, the bank account…

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.)\no y = 7,746.90(1.02)^x\no y = 7,746.90(0.02)^x\no y = 7,595.96(1.02)^x\no y = 7,595.96(0.02)^x

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.)\no y = 7,746.90(1.02)^x\no y = 7,746.90(0.02)^x\no y = 7,595.96(1.02)^x\no y = 7,595.96(0.02)^x

Answer

Answer:

C. $y = 7595.96(1.02)^x$

Explanation:

Step1: Identify compound - interest formula

The formula for compound - interest is $y = P(1 + r)^x$, where $P$ is the principal amount, $r$ is the annual interest rate as a decimal, and $x$ is the number of years.

Step2: Determine principal and rate

The initial amount (principal) $P=$7595.96$ and the rate of increase $r = 2%=0.02$.

Step3: Substitute values into formula

Substituting $P = 7595.96$ and $r = 0.02$ into $y = P(1 + r)^x$, we get $y=7595.96(1 + 0.02)^x=7595.96(1.02)^x$.