isabel deposits $6,000 into an account that earns 1.5% interest compounded monthly. assuming no more…

isabel deposits $6,000 into an account that earns 1.5% interest compounded monthly. assuming no more deposits and no withdrawals are made, how much money is in the account after 4 years? compound interest formula: $v(t)=p(1 + \frac{r}{n})^{nt}$\n$t$ = years since initial deposit\n$n$ = number of times compounded per year\n$r$ = annual interest rate (as a decimal)\n$p$ = initial (principal) investment\n$v(t)$ = value of investment after $t$ years\n$10,892.13\n$6,370.78\n$7,180.89\n$6,360.00

isabel deposits $6,000 into an account that earns 1.5% interest compounded monthly. assuming no more deposits and no withdrawals are made, how much money is in the account after 4 years? compound interest formula: $v(t)=p(1 + \frac{r}{n})^{nt}$\n$t$ = years since initial deposit\n$n$ = number of times compounded per year\n$r$ = annual interest rate (as a decimal)\n$p$ = initial (principal) investment\n$v(t)$ = value of investment after $t$ years\n$10,892.13\n$6,370.78\n$7,180.89\n$6,360.00

Answer

Explanation:

Step1: Identify the values

$P = 6000$, $r=0.015$, $n = 12$, $t = 4$

Step2: Substitute into formula

$V(t)=6000\left(1+\frac{0.015}{12}\right)^{12\times4}$

Step3: Calculate the exponent part

$12\times4 = 48$, $\frac{0.015}{12}=0.00125$, $1 + 0.00125=1.00125$

Step4: Calculate the power

$1.00125^{48}\approx1.061797$

Step5: Calculate the final value

$V(t)=6000\times1.061797 = 6370.782\approx6370.78$

Answer:

$6,370.78$