the principal represents an amount of money deposited in a savings account subject to compound interest at…

the principal represents an amount of money deposited in a savings account subject to compound interest at the given rate. answer parts (a) and (b). click the icon to view some finance formulas. a. find how much money there will be in the account after the given number of years. the amount of money in the account after 4 years is $ (round to the nearest hundredth as needed.) formulas in the provided formulas, a is the balance in the account after t years, p is the principal investment, r is the annual interest rate in decimal form, n is the number of compounding periods per year, and y is the investments effective annual yield in decimal form. $a = p(1+\frac{r}{n})^{nt}$ $p=\frac{a}{(1 + \frac{r}{n})^{nt}}$ $a = pe^{rt}$ $y=(1+\frac{r}{n})^{n}-1$ principal $5000 rate 1.5% compounded monthly time 4 years

the principal represents an amount of money deposited in a savings account subject to compound interest at the given rate. answer parts (a) and (b). click the icon to view some finance formulas. a. find how much money there will be in the account after the given number of years. the amount of money in the account after 4 years is $ (round to the nearest hundredth as needed.) formulas in the provided formulas, a is the balance in the account after t years, p is the principal investment, r is the annual interest rate in decimal form, n is the number of compounding periods per year, and y is the investments effective annual yield in decimal form. $a = p(1+\frac{r}{n})^{nt}$ $p=\frac{a}{(1 + \frac{r}{n})^{nt}}$ $a = pe^{rt}$ $y=(1+\frac{r}{n})^{n}-1$ principal $5000 rate 1.5% compounded monthly time 4 years

Answer

Explanation:

Step1: Identify the formula

We use the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of compounding periods per year, and $t$ is the number of years. Given $P=$5000$, $r = 0.015$ (since $1.5%=0.015$), $n = 12$ (compounded monthly), and $t = 4$.

Step2: Substitute the values into the formula

$A=5000(1 +\frac{0.015}{12})^{12\times4}$. First, calculate the value inside the parentheses: $\frac{0.015}{12}=0.00125$, then $1+\frac{0.015}{12}=1 + 0.00125=1.00125$. Next, calculate the exponent: $12\times4 = 48$. So, $A = 5000\times(1.00125)^{48}$.

Step3: Calculate $(1.00125)^{48}$

Using a calculator, $(1.00125)^{48}\approx1.061831$.

Step4: Calculate the value of $A$

$A=5000\times1.061831=$5309.16$.

Answer:

$5309.16$