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?\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\nt = years since initial deposit\nn = number of times compounded per year\nr = annual interest rate (as a decimal)\np = initial (principal) investment\nv(t) = value of investment after t years\n\n$6,360.00\n$6,370.78\n$7,180.89\n$10,892.13

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?\ncompound interest formula: $v(t)=p(1 + \\frac{r}{n})^{nt}$\nt = years since initial deposit\nn = number of times compounded per year\nr = annual interest rate (as a decimal)\np = initial (principal) investment\nv(t) = value of investment after t years\n\n$6,360.00\n$6,370.78\n$7,180.89\n$10,892.13

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

First, $\frac{0.015}{12}=0.00125$, then $12\times4 = 48$, and $1 + 0.00125=1.00125$. So we have $V(t)=6000\times(1.00125)^{48}$

Step4: Calculate the power value

$(1.00125)^{48}\approx1.061297$

Step5: Calculate the final value

$V(t)=6000\times1.061297 = 6370.782\approx6370.78$

Answer:

$6370.78$ (corresponding to the option: $$6,370.78$)