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
\(t=12\ years\\ P=\$9,350\\ r=0.0258\
=365\)
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
\(t=12\ years\\ A=\$10,891.31\\ r=0.029\
=4\)
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 \(\$225\)
