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steposvetlana [31]
4 years ago
6

Juno deposited $750 in a savings account that earns 4% interest compounded annually. If she does not deposit or withdraw any mor

e money, how much money will there be in the account after 13 years? Round your answer to the nearest dollar. A. $1249 B. $1360 C. $1427 D. $1140
Mathematics
2 answers:
TiliK225 [7]4 years ago
6 0
<span>The correct answer is A) $1249.

Explanation:
The amount of interest is calculated using the formula
</span>A=p(1+\frac{r}{n})^{nt},
<span>
where a is the amount of principal, r is the interest rate as a decimal number, n is the number of times per year the interest is compounded, and t is the number of years.

Our principal is $750, the interest rate is 4%=4/100=0.04, n is 1, and t is 13:
</span>A=750(1+\frac{0.04}{1})^{13\times1}=750(1+0.04)^{13}=750(1.04)^{13}=1248.81<span>,

which rounds to 1249.</span>
aniked [119]4 years ago
3 0

Answer:

A. $1249.

Step-by-step explanation:

We have been given that Juno deposited $750 in a savings account that earns 4% interest compounded annually. We are asked to find the amount of money in the account after 13 years.

To solve our given problem, we will use compound interest formula. A=P(1+\frac{r}{n})^{nt}, where,

A = Amount after t years,

P = Principal amount,

r = Interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

Let us convert our given interest rate in decimal form.

4\%=\frac{4}{100}=0.04

Upon substituting our given values in compound interest formula we will get,

A=\$750(1+\frac{0.04}{1})^{1\cdot 13}

A=\$750(1+0.04)^{13}  

A=\$750(1.04)^{13}  

A=\$750\times 1.665073507310388  

A=\$1248.805130482791\approx \$1249  

Therefore, there will be $1249 in Juno's account after 13 years and option A is the correct choice.

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Using a system of equations, it is found that Peter had $48 at first.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

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