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Sagot :
Suzanne deposits $3000 in an account that earns simple interest at an annual rate of 4.6%. Derek deposits $3000 in an account that eams compound interest at an annual rate of 4.6% and is compounded annually
step 1
simple interest
we have that
The simple interest formula is equal to
[tex]A=P(1+rt)[/tex]
[tex]A=P\mleft(1+rt\mright)[/tex]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
P=$3,000
r=4.6%=0.046
t=1 year
substitute in the formula
[tex]\begin{gathered} A=3,000(1+0.046\cdot1) \\ A=\$3,138 \end{gathered}[/tex]therefore
Suzanne's annual interest is (3,138-3,000)=$138
Suzanne's balance is $3,138
step 2
compound interest
we have that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[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
P=$3,000
r=4.6%=0.046
t=1 year
n=1
substitute
[tex]\begin{gathered} A=3,000(1+\frac{0.046}{1})^1 \\ A=\$3,138 \end{gathered}[/tex]therefore
Derek's annual interest is (3,138-3,000)=$138
Dereck's balance is $3,138
Is the same
For two years
t=2 years
simple interest
[tex]\begin{gathered} A=3,000(1+0.046\cdot2) \\ A=\$3,276 \end{gathered}[/tex]Suzanne's annual interest is (3,276-3,000)=$276
Suzanne's balance is $3,276
Compound interest
[tex]\begin{gathered} A=3,000(1+\frac{0.046}{1})^2 \\ A=\$3,282.35 \end{gathered}[/tex]interest is $282,35
Balance is $ 3,282.35
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