Respuesta :
part A)
[tex]\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\ \left. \qquad \qquad \right.(\textit{payments at the end of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right][/tex]
[tex]\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array} \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &1200\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &12 \end{cases}[/tex]
[tex]\bf A=1200\left[ \cfrac{\left( 1+\frac{0.05}{1} \right)^{1\cdot 12}-1}{\frac{0.05}{1}} \right]\implies A\approx 19100.55[/tex]
part B)
so, for the next 11 years, she didn't make any deposits on it and simple let it sit and collect interest, compounded annually at 5%.
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \ \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$19100.55\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &11 \end{cases} \\\\\\ A=19100.55\left(1+\frac{0.05}{1}\right)^{1\cdot 11}\implies A\approx 32668.42[/tex]
part C)
well, for 12 years she deposited 1200 bucks, that means 12 * 1200, or 14,400.
now, here she is, 12+11, or 23 years later, and she's got 32,668.42 bucks?
all that came out of her pocket was 14,400, so 32,668.42 - 14,400, is how much she earned in interest.
[tex]\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\ \left. \qquad \qquad \right.(\textit{payments at the end of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right][/tex]
[tex]\bf \qquad \begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array} \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &1200\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &12 \end{cases}[/tex]
[tex]\bf A=1200\left[ \cfrac{\left( 1+\frac{0.05}{1} \right)^{1\cdot 12}-1}{\frac{0.05}{1}} \right]\implies A\approx 19100.55[/tex]
part B)
so, for the next 11 years, she didn't make any deposits on it and simple let it sit and collect interest, compounded annually at 5%.
[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \ \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$19100.55\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &11 \end{cases} \\\\\\ A=19100.55\left(1+\frac{0.05}{1}\right)^{1\cdot 11}\implies A\approx 32668.42[/tex]
part C)
well, for 12 years she deposited 1200 bucks, that means 12 * 1200, or 14,400.
now, here she is, 12+11, or 23 years later, and she's got 32,668.42 bucks?
all that came out of her pocket was 14,400, so 32,668.42 - 14,400, is how much she earned in interest.
a. The value of the Ira at the end of 12 years is $19,100.55.
b. The value of the investment at the end of the next 11 years is $32,668.43.
c. Interest earn is $18,288.43.
a. Using this formula to determine the value of the Ira at the end of 12 years
A=Pmt [(1+r)^n-1]/r
Let plug in the formula
A=1,200[(1+0.05)^12-1]/0.05
A=1,200[(1.05)^12-1]/0.05
A=$1,200(0.795856)/0.05
A=$955.02759/0.05
A=$19,100.55
b. The value of the Ira at the end of 11 years is:
$19,100.55(1+0.05)^11
=$19,100.55(1.05)^11
=$19,100.55(1.710339358)
=$32,668.43
c. Interest earn
Total investment=1200(12)
Total investment= $14,400
Now let calculate the interest earned
Interest earned = $32,668.43 - $14,400
Interest earned= $18,288.43
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