finance. a person wishes to have $15,800 cash for a new car 4 years from now. how much should be placed in…

finance. a person wishes to have $15,800 cash for a new car 4 years from now. how much should be placed in an account now, if the account pays 6.2% annual interest rate, compounded weekly?\n$ (round to the nearest dollar.)

finance. a person wishes to have $15,800 cash for a new car 4 years from now. how much should be placed in an account now, if the account pays 6.2% annual interest rate, compounded weekly?\n$ (round to the nearest dollar.)

Answer

Explanation:

Step1: Identify the 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 (the initial amount of money), $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the number of years. We want to find $P$, so we can rewrite the formula as $P=\frac{A}{(1 +\frac{r}{n})^{nt}}$.

Step2: Convert the given values to the appropriate form

The future value $A=$15800$, the annual interest rate $r = 6.2%=0.062$, since interest is compounded weekly, $n = 52$, and the number of years $t = 4$.

Step3: Substitute the values into the formula

$P=\frac{15800}{(1+\frac{0.062}{52})^{52\times4}}$. First, calculate the value inside the parentheses: $\frac{0.062}{52}\approx0.0011923$, then $1+\frac{0.062}{52}=1 + 0.0011923=1.0011923$. Next, calculate the exponent: $52\times4 = 208$. Then $(1.0011923)^{208}\approx1.28177$. Finally, $P=\frac{15800}{1.28177}\approx12326$.

Answer:

$12326$