Answer:
Step-by-step explanation:
[tex]A=P(1+\frac{r}{n})^{nt[/tex]
A = future value ($)
P = principal (orignal amount - $400)
R = rate (written as a decimal - 3%/100= 0.03)
N = compound (yearly - 1)
T = number of years (0, 5, 10, 15)
Now, substitute each number into the formula and record them in the table.
0 years:
[tex]A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*0}\\\\A=400(1+\frac{0.03}{1})^0 < ==any\ number\ raised\ to\ zero=1\\\\A=400(1)\\\\A=\$400[/tex]
5 years:
[tex]A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*5}\\\\A=400(1+\frac{0.03}{1})^{5}\\\\A=400(1+{0.03})^{5}\\\\A=400({2.03})^{5}\\\\A=400({1.1593})\\\\A=\$463.72 < ==round\ to\ the\ nearest\ dollar\\\\A=\$464[/tex]
10 years:
[tex]A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*10}\\\\A=400(1+\frac{0.03}{1})^{10}\\\\A=400(1+{0.03})^{10}\\\\A=400({1.03})^{10}\\\\A=400({1.3439})\\\\A=\$537.56 < ==round\ to\ the\ nearest\ dollar\\\\A=\$538[/tex]
15 years:
[tex]A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*15}\\\\A=400(1+\frac{0.03}{1})^{15}\\\\A=400(1+{0.03})^{15}\\\\A=400({1.03})^{15}\\\\A=400({1.5579})\\\\A=623.16 < ==round\ to\ the\ nearest\ dollar\\\\A=\$623[/tex]
20 years:
[tex]A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*20}\\\\A=400(1+\frac{0.03}{1})^{20}\\\\A=400(1+{0.03})^{20}\\\\A=400({1.03})^{20}\\\\A=400({1.8061})\\\\A=722.44 < ==round\ to\ the\ nearest\ dollar\\\\A=\$722[/tex]
Note that since we are rounding, answers may vary by a dollar or two.
Hope this helps!
