Question

At the beginning of year 1, Josie invests $400 at an annual compound interest

Answer 1

Answer:

The explicit formula that can be used is $A=\ 400(1.05)^{2}$

The account's balance at the beginning of year 3 is $A=\ 441$  

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

$t=2\ years\\ P=\ 400\\ r=0.05\\n=1$  

substitute in the formula above  

$A=\ 400(1+\frac{0.05}{1})^{1*2}$  

$A=\ 400(1.05)^{2}$

$A=\ 441$