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Sagot :
Answer:
annualy=$3689.62
semiannually=$3695.27
monthly=$3700.06
weekly=$3700.81
daily=$3701.00
Continuously=$3701.03
Step-by-step explanation:
Given:
P=3000
r=3%
t=7 years
Formula used:
Where,
A represents Accumulated amount
P represents (or) invested amount
r represents interest rate
t represents time in years
n represents accumulated or compounded number of times per year
Solution:
(i)annually
n=1 time per year
[tex]A=3000[1+\frac{0.03}{1} ]^1^(^7^)\\ =3000(1.03)^7\\ =3689.621596\\[/tex]
On approximating the values,
A=$3689.62
(ii)semiannually
n=2 times per year
[tex]A=3000[1+\frac{0.03}{2}^{2(4)} ]\\ =3000[1+0.815]^14\\ =3695.267192[/tex]
On approximating the values,
A=$3695.27
(iii)monthly
n=12 times per year
[tex]A=3000[1+\frac{0.03}{12}^{12(7)} \\ =3000[1+0.0025]^84\\ =3700.0644[/tex]
On approximating,
A=$3700.06
(iv) weekly
n=52 times per year
[tex]A=3000[1+\frac{0.03}{52}]^3^6 \\ =3000(1.23360336)\\ =3700.81003[/tex]
On approximating,
A=$3700.81
(v) daily
n=365 time per year
[tex]A=3000[1+\frac{0.03}{365}]^{365(7)} \\ =3000[1.000082192]^{2555}\\ =3701.002234[/tex]
On approximating the values,
A=$3701.00
(vi) Continuously
[tex]A=Pe^r^t\\ =3000e^{\frac{0.03}{1}(7) }\\ =3000e^{0.21} \\ =3000(1.23367806)\\ =3701.03418\\[/tex]
On approximating the value,
A=$3701.03
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