Answer :
We begin with the formula for compound interest, which in this case is given by
[tex]$$
A(t) = 60000 (1.06)^t,
$$[/tex]
where:
- [tex]$60000$[/tex] is the initial amount (the principal),
- [tex]$1.06$[/tex] is the growth factor corresponding to a [tex]$6\%$[/tex] interest rate,
- [tex]$t$[/tex] is the time in years.
We now calculate [tex]$A(t)$[/tex] for the specified values of [tex]$t$[/tex].
1. For [tex]$t = 0$[/tex]:
[tex]$$
A(0) = 60000 (1.06)^0 = 60000 \cdot 1 = 60000.00.
$$[/tex]
2. For [tex]$t = 1$[/tex]:
[tex]$$
A(1) = 60000 (1.06)^1 = 60000 \cdot 1.06 = 63600.00.
$$[/tex]
3. For [tex]$t = 2$[/tex]:
[tex]$$
A(2) = 60000 (1.06)^2.
$$[/tex]
Evaluating [tex]$(1.06)^2$[/tex] yields approximately [tex]$1.1236$[/tex], so
[tex]$$
A(2) \approx 60000 \cdot 1.1236 = 67416.00.
$$[/tex]
4. For [tex]$t = 4$[/tex]:
[tex]$$
A(4) = 60000 (1.06)^4.
$$[/tex]
This value approximates to
[tex]$$
A(4) \approx 75748.62.
$$[/tex]
5. For [tex]$t = 8$[/tex]:
[tex]$$
A(8) = 60000 (1.06)^8.
$$[/tex]
The approximate value is
[tex]$$
A(8) \approx 95630.88.
$$[/tex]
6. For [tex]$t = 10$[/tex]:
[tex]$$
A(10) = 60000 (1.06)^{10}.
$$[/tex]
Which gives approximately
[tex]$$
A(10) \approx 107450.86.
$$[/tex]
7. For [tex]$t = 20$[/tex]:
[tex]$$
A(20) = 60000 (1.06)^{20}.
$$[/tex]
This computation results in
[tex]$$
A(20) \approx 192428.13.
$$[/tex]
Thus, the completed table of function values is:
[tex]$$
\begin{array}{|c|c|}
\hline
t & A(t) \\
\hline
0 & 60000.00 \\
\hline
1 & 63600.00 \\
\hline
2 & 67416.00 \\
\hline
4 & 75748.62 \\
\hline
8 & 95630.88 \\
\hline
10 & 107450.86 \\
\hline
20 & 192428.13 \\
\hline
\end{array}
$$[/tex]
Each amount has been rounded to two decimal places as required.
[tex]$$
A(t) = 60000 (1.06)^t,
$$[/tex]
where:
- [tex]$60000$[/tex] is the initial amount (the principal),
- [tex]$1.06$[/tex] is the growth factor corresponding to a [tex]$6\%$[/tex] interest rate,
- [tex]$t$[/tex] is the time in years.
We now calculate [tex]$A(t)$[/tex] for the specified values of [tex]$t$[/tex].
1. For [tex]$t = 0$[/tex]:
[tex]$$
A(0) = 60000 (1.06)^0 = 60000 \cdot 1 = 60000.00.
$$[/tex]
2. For [tex]$t = 1$[/tex]:
[tex]$$
A(1) = 60000 (1.06)^1 = 60000 \cdot 1.06 = 63600.00.
$$[/tex]
3. For [tex]$t = 2$[/tex]:
[tex]$$
A(2) = 60000 (1.06)^2.
$$[/tex]
Evaluating [tex]$(1.06)^2$[/tex] yields approximately [tex]$1.1236$[/tex], so
[tex]$$
A(2) \approx 60000 \cdot 1.1236 = 67416.00.
$$[/tex]
4. For [tex]$t = 4$[/tex]:
[tex]$$
A(4) = 60000 (1.06)^4.
$$[/tex]
This value approximates to
[tex]$$
A(4) \approx 75748.62.
$$[/tex]
5. For [tex]$t = 8$[/tex]:
[tex]$$
A(8) = 60000 (1.06)^8.
$$[/tex]
The approximate value is
[tex]$$
A(8) \approx 95630.88.
$$[/tex]
6. For [tex]$t = 10$[/tex]:
[tex]$$
A(10) = 60000 (1.06)^{10}.
$$[/tex]
Which gives approximately
[tex]$$
A(10) \approx 107450.86.
$$[/tex]
7. For [tex]$t = 20$[/tex]:
[tex]$$
A(20) = 60000 (1.06)^{20}.
$$[/tex]
This computation results in
[tex]$$
A(20) \approx 192428.13.
$$[/tex]
Thus, the completed table of function values is:
[tex]$$
\begin{array}{|c|c|}
\hline
t & A(t) \\
\hline
0 & 60000.00 \\
\hline
1 & 63600.00 \\
\hline
2 & 67416.00 \\
\hline
4 & 75748.62 \\
\hline
8 & 95630.88 \\
\hline
10 & 107450.86 \\
\hline
20 & 192428.13 \\
\hline
\end{array}
$$[/tex]
Each amount has been rounded to two decimal places as required.
