practice assignment 6.1 exponential functio\n$6000 are invested in a bank account at an interest rate of 7…

Explanation:

Step1: Identify the compound - interest formulas

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $n$ is the number of times interest is compounded per year, $t$ is the number of years, and $A$ is the amount of money accumulated after $n$ years, including interest. For continuous compounding, the formula is $A = Pe^{rt}$. Given $P=$6000$, $r = 0.07$, and $t = 11$.

Step2: Calculate when compounded annually ($n = 1$)

Substitute $P = 6000$, $r=0.07$, $n = 1$, and $t = 11$ into the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$. $A=6000(1 +\frac{0.07}{1})^{1\times11}=6000(1.07)^{11}$ $A\approx6000\times2.104851709\approx12629.11$

Step3: Calculate when compounded quarterly ($n = 4$)

Substitute $P = 6000$, $r = 0.07$, $n = 4$, and $t = 11$ into the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$. $A=6000(1+\frac{0.07}{4})^{4\times11}=6000(1 + 0.0175)^{44}$ $A=6000\times(1.0175)^{44}$ Using a calculator, $(1.0175)^{44}\approx2.149778777$, so $A\approx6000\times2.149778777\approx12898.67$

Step4: Calculate when compounded monthly ($n = 12$)

Substitute $P = 6000$, $r = 0.07$, $n = 12$, and $t = 11$ into the compound - interest formula $A = P(1+\frac{r}{n})^{nt}$. $A=6000(1+\frac{0.07}{12})^{12\times11}=6000(1+\frac{0.07}{12})^{132}$ $A = 6000\times(1+\frac{0.07}{12})^{132}$ Using a calculator, $(1+\frac{0.07}{12})^{132}\approx2.157857777$, so $A\approx6000\times2.157857777\approx12947.15$

Step5: Calculate when compounded continuously

Substitute $P = 6000$, $r = 0.07$, and $t = 11$ into the continuous - compounding formula $A = Pe^{rt}$. $A=6000e^{0.07\times11}=6000e^{0.77}$ Using a calculator, $e^{0.77}\approx2.16091636$, so $A\approx6000\times2.16091636\approx12965.50$

Answer:

$12629.11$ $12898.67$ $12947.15$ $12965.50$