Respuesta :
Let:
P be the initial amount of money called the Principal,
compounded n times a year, with an r annual interest rate, then after
t many years, the amount of money A is given by the formula:
[tex]A=P(1+ \frac{r}{n} )^{nt} [/tex]
Remark
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r is generally a percentage like 3%, 7% etc and are applied in the formula as 0.03, 0.07...,
the interest is compounded generally annually (n=1), quarterly (n=4), monthly (n=12), etc...
t is in years,
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Thus, in our problem, P=$4,000, r=4%=0.04, n=12, t=1
Applying the formula:
[tex]A=P(1+ \frac{r}{n} )^{nt} [/tex]
[tex]A=4,000* (1+ \frac{0.04}{12} )^{12*1}=4,000(1+0.0033)^{12} [/tex]
[tex]=4,000*1.0407=4162.97[/tex]
At the end of the year Janise has 4,162.97-4,000≈163 more dollars than the Principal amount.
Thus, the interest earned during the year is 163 $.
Answer: $163
P be the initial amount of money called the Principal,
compounded n times a year, with an r annual interest rate, then after
t many years, the amount of money A is given by the formula:
[tex]A=P(1+ \frac{r}{n} )^{nt} [/tex]
Remark
----------------------------------------------------------------------------------
r is generally a percentage like 3%, 7% etc and are applied in the formula as 0.03, 0.07...,
the interest is compounded generally annually (n=1), quarterly (n=4), monthly (n=12), etc...
t is in years,
-------------------------------------------------------------------------------------
Thus, in our problem, P=$4,000, r=4%=0.04, n=12, t=1
Applying the formula:
[tex]A=P(1+ \frac{r}{n} )^{nt} [/tex]
[tex]A=4,000* (1+ \frac{0.04}{12} )^{12*1}=4,000(1+0.0033)^{12} [/tex]
[tex]=4,000*1.0407=4162.97[/tex]
At the end of the year Janise has 4,162.97-4,000≈163 more dollars than the Principal amount.
Thus, the interest earned during the year is 163 $.
Answer: $163
Answer:
$162.97 is the exact answer.
Step-by-step explanation:
