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The amount of money, [tex]$A$[/tex], accrued at the end of [tex]$n$[/tex] years when a certain amount, [tex]$P$[/tex], is invested at a compound annual rate, [tex]$r$[/tex], is given by [tex]A = P(1 + r)^n[/tex].

If a person invests [tex]$\$[/tex]300[tex]$ in an account that pays $[/tex]8.5\%[tex]$ interest compounded annually, find the balance after 15 years.

A. $[/tex]\[tex]$940.02$[/tex]
B. [tex]$\$[/tex]1106.62[tex]$
C. $[/tex]\[tex]$1019.92$[/tex]
D. [tex]$\$[/tex]4882.50$



Answer :

GinnyAnswer
To solve for the balance after 15 years when an initial principal is invested with compound interest, we'll use the compound interest formula:

[tex]\[ A = P(1 + r)^n \][/tex]

where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( n \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form).
- [tex]\( n \)[/tex] is the number of years the money is invested.

Given the values:
- [tex]\( P = \$ 300 \)[/tex]
- [tex]\( r = 8.5\% = 0.085 \)[/tex]
- [tex]\( n = 15 \)[/tex] years

We plug these values into the compound interest formula:

[tex]\[ A = 300 (1 + 0.085)^{15} \][/tex]

First, calculate [tex]\( 1 + r \)[/tex]:
[tex]\[ 1 + 0.085 = 1.085 \][/tex]

Next, raise this base to the power of [tex]\( n \)[/tex]:
[tex]\[ 1.085^{15} \approx 3.399742 \][/tex]

Now, multiply this result by [tex]\( P \)[/tex] to find [tex]\( A \)[/tex]:
[tex]\[ A = 300 \times 3.399742 = 1019.9228636463336 \][/tex]

Therefore, the balance after 15 years is approximately:
[tex]\[ \$ 1019.92 \][/tex]

Among the given options, the correct answer is:
[tex]\[ \$ 1019.92 \][/tex]

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