Question

HELP PLEASE!!!

Answer 1

1-(b)
2-(b)111600
3-(a)
4-(d)

Answer 2

Answer:

Part 1) Option B. $\ 3,392.79$

Part 2) Option C. $\ 3,100$

Part 3) Option A. $\ 7,700$  

Part 4) Option D. About $\ 225$

Step-by-step explanation:

Part 1) 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  

$t=12\ years\\ P=\ 9,350\\ r=0.0258\\n=365$  

substitute in the formula above  

$A=\ 9,350(1+\frac{0.0258}{365})^{365*12}=\ 12,742.79$  

Find the interest

$I=A-P=\ 12,742.79-\ 9,350=\ 3,392.79$

Part 2) we know that

The simple interest formula is equal to

$I=P(rt)$

$P=I/(rt)$

where

I is the interest amount  

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

$t=6\ years\\ I=\ 334.80\\ P=?\\r=0.018$

substitute in the formula above

$P=\ 334.80/(0.018*6)=\ 3,100$

Part 3) we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

$P=A/[(1+\frac{r}{n})^{nt}]$  

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  

$t=12\ years\\ A=\ 10,891.31\\ r=0.029\\n=4$  

substitute in the formula above  

$P=\ 10,891.31/[(1+\frac{0.029}{4})^{4*12}]=\ 7,700$  

Part 4)

The account with compound interest earn about $\ 760$

The account with simple interest earn about $\ 520$

see the attached figure

The difference is equal to

 $\ 760-\ 520=\ 240$

therefore

the answer is about $\ 225$