Answer:
Part 1) Option B. [tex]\$3,392.79[/tex]
Part 2) Option C. [tex]\$3,100[/tex]
Part 3) Option A. [tex]\$7,700[/tex]
Part 4) Option D. About [tex]\$225[/tex]
Step-by-step explanation:
Part 1) we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
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
[tex]t=12\ years\\ P=\$9,350\\ r=0.0258\\n=365[/tex]
substitute in the formula above
[tex]A=\$9,350(1+\frac{0.0258}{365})^{365*12}=\$12,742.79[/tex]
Find the interest
[tex]I=A-P=\$12,742.79-\$9,350=\$3,392.79[/tex]
Part 2) we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
[tex]P=I/(rt)[/tex]
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
[tex]t=6\ years\\ I=\$334.80\\ P=?\\r=0.018[/tex]
substitute in the formula above
[tex]P=\$334.80/(0.018*6)=\$3,100[/tex]
Part 3) we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[tex]P=A/[(1+\frac{r}{n})^{nt}][/tex]
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
[tex]t=12\ years\\ A=\$10,891.31\\ r=0.029\\n=4[/tex]
substitute in the formula above
[tex]P=\$10,891.31/[(1+\frac{0.029}{4})^{4*12}]=\$7,700[/tex]
Part 4)
The account with compound interest earn about [tex]\$760[/tex]
The account with simple interest earn about [tex]\$520[/tex]
see the attached figure
The difference is equal to
[tex]\$760-\$520=\$240[/tex]
therefore
the answer is about [tex]\$225[/tex]
