Question

4400 dollars is placed in an account with an annual interest rate of 8.25%. How much
will be in the account after 29 years, to the nearest cent?

Answer 1

Answer:

43839.13

Step-by-step explanation:

\text{Exponential Functions:}

Exponential Functions:

y=ab^x

y=ab

x

a=\text{starting value = }4400

a=starting value = 4400

r=\text{rate = }8.25\% = 0.0825

r=rate = 8.25%=0.0825

\text{Exponential Growth:}

Exponential Growth:

b=1+r=1+0.0825=1.0825

b=1+r=1+0.0825=1.0825

\text{Write Exponential Function:}

Write Exponential Function:

y=4400(1.0825)^x

y=4400(1.0825)

x

Put it all together

\text{Plug in time for x:}

Plug in time for x:

y=4400(1.0825)^{29}

y=4400(1.0825)

29

y= 43839.1295315

y=43839.1295315

Evaluate

y\approx 43839.13

y≈43839.13

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