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a $1,600 principal earns 7% annual interest, compounded semiannually (twice per year). After 33 years, what is the balance in the account?

Respuesta :

aachen

Given is the Principal amount, P = 1600 dollars.

Given the Annual interest is 7% i.e. r = 0.07

Given the Compounding period is semi-annually i.e. n = 2.

Given is the Time of investment, t = 33 years.

It says to find the Final Value of invested amount in the account after 33 years.

We know the formula for Future Value of Money is given as follows :-

[tex] Future \;\;Value = P*(1+\frac{r}{n})^{nt} \\\\
Future \;\;Value = 1600*(1+\frac{0.07}{2})^{(2*33)} \\\\
Future \;\;Value = 1600*(1+0.035)^{66} \\\\
Future \;\;Value = 1600*(1.035)^{66} \\\\
Future \;\;Value = 1600*(9.684185201) \\\\
Future \;\;Value = 15494.69632 \\\\
Future \;\;Value = 15,494.70 \;\;dollars [/tex]

Hence, the final balance would be 15,494.70 dollars.

Answer:

After 33 years balance in the account  = A= $15494.696

Explanation:

We will applying the compound interest formula.

[tex] A =  P(1 +\frac{r}{n})^{nt} [/tex]  

Where,

A = the future value of the investment/loan, including interest

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per year

t = the number of years

Given that,

P = $1,600

r = 7%  =  [tex] \frac{7}{100}  [/tex] = 0.07

n = 2 (because of twice in a year)

t = 33 years

A= [tex] 1600(1 + \frac{0.035}{2}) ^{2*33}  [/tex] 

A =[tex]  1600 (1.035)^{66}  [/tex] 

A= $15494.696


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