Answer:
(b)
[tex]t\approx4.11\text{ years}[/tex]
(c) 14 years
(d) At t = 10 years, S = $18,123.77
At t = 20, S = $36,496.80
Explanation:
b) We're asked to use the below graph to estimate when the future value will be $12,000;
Notice that each interval on the vertical axis represents 4,000 units and each interval on the horizontal axis represents 2 units.
Looking at the given graph, we can see that at S = $12,000 on the vertical axis, we can read off the intersection point on the horizontal axis from the graph and it is at t equals to approximately 4.11 years.
So the correction option is;
[tex]t\approx4.11\text{years}[/tex]
c) When S = $23,980, let's go ahead and determine the value of t by substituting S with 23,980 and solving for t as seen below;
[tex]\begin{gathered} S=9000e^{0.07t} \\ 23980.11=9000e^{0.07t} \\ \frac{23980.11}{9000}=e^{0.07t} \\ 2.6645=e^{0.07t} \\ \ln2.6645=lne^{0.07t} \\ \ln2.6645=0.07t \\ t=\frac{2.6645}{0.07} \\ t=14\text{ years} \end{gathered}[/tex]
Therefore, at S = $23.980.11, t = 14 years
d) When t = 10, let's go ahead and solve for S as seen below;
[tex]\begin{gathered} S=9000e^{0.07t} \\ When\text{ t = 10,} \\ S=9000e^{0.07*10} \\ S=9000e^{0.7} \\ S=9000\left(2.0138\right) \\ S=18,123.77 \end{gathered}[/tex]
So at t = 10 years, S = $18,123.77
At t = 20, let's solve for S;
[tex]\begin{gathered} S=9000e^{0.07t} \\ when\text{ t = 20} \\ S=9000e^{0.07*20} \\ S=9000e^{1.4} \\ S=9000\left(4.0552\right) \\ S=36,496.80 \end{gathered}[/tex]
Therefore, at t = 20, S = $36,496.80
