finance. a person wishes to have $17,600 cash for a new car 3 years from now. how much should be placed in…

finance. a person wishes to have $17,600 cash for a new car 3 years from now. how much should be placed in an account now, if the account pays 5.6% annual interest rate, compounded weekly? $ (round to the nearest dollar.)

finance. a person wishes to have $17,600 cash for a new car 3 years from now. how much should be placed in an account now, if the account pays 5.6% annual interest rate, compounded weekly? $ (round to the nearest dollar.)

Answer

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the future value, $P$ is the principal amount (initial deposit), $r$ is the annual interest rate (in decimal form), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. We want to solve for $P$, so we can rewrite the formula as $P=\frac{A}{(1 +\frac{r}{n})^{nt}}$.

Step2: Convert values to appropriate form

The annual interest rate $r = 5.6%=0.056$, the number of years $t = 3$, the future value $A=$17600$. Since interest is compounded weekly, $n = 52$.

Step3: Substitute values into the formula

$P=\frac{17600}{(1+\frac{0.056}{52})^{52\times3}}$. First, calculate the value inside the parentheses: $\frac{0.056}{52}\approx0.001077$, then $1+\frac{0.056}{52}=1 + 0.001077=1.001077$. Next, calculate the exponent: $52\times3 = 156$. Then, $(1.001077)^{156}\approx1.18277$. Finally, $P=\frac{17600}{1.18277}\approx14889$.

Answer:

$14889$