14. -/1 points 0/12 submissions used\nan investment of $6,000 is deposited into an account in which interest…

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula when compounded monthly is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. Here, $P=$6000$, $r = 0.06$, and $n = 12$.

Step2: Calculate the amount for $t = 1$ year

Substitute $t = 1$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times1}\ &=6000(1 + 0.005)^{12}\ &=6000\times(1.005)^{12}\ &\approx6000\times1.0616778\ &\approx$6370.07 \end{align*} ]

Step3: Calculate the amount for $t = 2$ years

Substitute $t = 2$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times2}\ &=6000(1.005)^{24}\ &\approx6000\times1.1271602\ &\approx$6762.96 \end{align*} ]

Step4: Calculate the amount for $t = 3$ years

Substitute $t = 3$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times3}\ &=6000(1.005)^{36}\ &\approx6000\times1.1966806\ &\approx$7180.08 \end{align*} ]

Step5: Calculate the amount for $t = 4$ years

Substitute $t = 4$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times4}\ &=6000(1.005)^{48}\ &\approx6000\times1.2704892\ &\approx$7622.93 \end{align*} ]

Step6: Calculate the amount for $t = 5$ years

Substitute $t = 5$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times5}\ &=6000(1.005)^{60}\ &\approx6000\times1.3488502\ &\approx$8093.10 \end{align*} ]

Step7: Calculate the amount for $t = 6$ years

Substitute $t = 6$ into the formula: [ \begin{align*} A&=6000\left(1+\frac{0.06}{12}\right)^{12\times6}\ &=6000(1.005)^{72}\ &\approx6000\times1.4320449\ &\approx$8592.27 \end{align*} ]

Answer:

Time (years)	Amount
1	$$6370.07$
2	$$6762.96$
3	$$7180.08$
4	$$7622.93$
5	$$8093.10$
6	$$8592.27$