Question

I need help with D, i’m confused on it.

Answer 1

Answer:

Step-by-step explanation:

$A=P(1+\frac{r}{n})^{nt$

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:

$A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*0}\\\\A=400(1+\frac{0.03}{1})^0 < ==any\ number\ raised\ to\ zero=1\\\\A=400(1)\\\\A=\ 400$

5 years:

$A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*5}\\\\A=400(1+\frac{0.03}{1})^{5}\\\\A=400(1+{0.03})^{5}\\\\A=400({2.03})^{5}\\\\A=400({1.1593})\\\\A=\ 463.72 < ==round\ to\ the\ nearest\ dollar\\\\A=\ 464$

10 years:

$A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*10}\\\\A=400(1+\frac{0.03}{1})^{10}\\\\A=400(1+{0.03})^{10}\\\\A=400({1.03})^{10}\\\\A=400({1.3439})\\\\A=\ 537.56 < ==round\ to\ the\ nearest\ dollar\\\\A=\ 538$

15 years:

$A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*15}\\\\A=400(1+\frac{0.03}{1})^{15}\\\\A=400(1+{0.03})^{15}\\\\A=400({1.03})^{15}\\\\A=400({1.5579})\\\\A=623.16 < ==round\ to\ the\ nearest\ dollar\\\\A=\ 623$

20 years:

 $A=P(1+\frac{r}{n})^{nt}\\\\A=400(1+\frac{0.03}{1})^{1*20}\\\\A=400(1+\frac{0.03}{1})^{20}\\\\A=400(1+{0.03})^{20}\\\\A=400({1.03})^{20}\\\\A=400({1.8061})\\\\A=722.44 < ==round\ to\ the\ nearest\ dollar\\\\A=\ 722$

Note that since we are rounding, answers may vary by a dollar or two.

Hope this helps!