Question

Gabriel deposits $2,500 into each of two savings accounts.

Answer 1

Answer:

Option B) $5,612.16

Step-by-step explanation:

Part 1) Account I earns 4% annual simple interest.

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

$t=3\ years\\ P=\ 2,500\\r=4\%=4/100=0.04$

substitute in the formula above

$A=2,500(1+0.04*3)$

$A=2,500(1.12)$

$A=\ 2,800$

Part 2) Account II earns 4% interest compounded annually.

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  

$t=3\ years\\ P=\ 2,500\\r=4\%=4/100=0.04\\n=1$  

substitute in the formula above  

$A=2,500*(1+\frac{0.04}{1})^{3}$

$A=2,500*(1.04)^{3}$

$A=\ 2,812.16$

Part 3) What is the sum of the balances of Account I and Account II at the end of 3 years?

Sum the two final investment

$\ 2,800+\ 2,812.16=\ 5.612.16$