If $3000 is invested at 3% interest, find the value of the investment at the end of 7 years if the interest is compounded as follows. (Round

Question

If $3000 is invested at 3% interest, find the value of the investment at the end of 7 years if the interest is compounded as follows. (Round your answers to the nearest cent.)
(i) annually
(ii) semiannually
(iii) monthly
(iv) weekly
(v) daily
(vi) continuously

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King 3 years 2021-07-30T23:19:53+00:00 1 Answers 10 views 0

Answers ( )

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    2021-07-30T23:20:58+00:00

    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

    A=3000[1+\frac{0.03}{1} ]^1^(^7^)\\  =3000(1.03)^7\\  =3689.621596\\

    On approximating the values,

    A=$3689.62

    (ii)semiannually

    n=2 times per year

    A=3000[1+\frac{0.03}{2}^{2(4)} ]\\   =3000[1+0.815]^14\\   =3695.267192

    On approximating the values,

    A=$3695.27

    (iii)monthly

    n=12 times per year

    A=3000[1+\frac{0.03}{12}^{12(7)} \\   =3000[1+0.0025]^84\\   =3700.0644

    On approximating,

    A=$3700.06

    (iv) weekly

    n=52 times per year

    A=3000[1+\frac{0.03}{52}]^3^6 \\   =3000(1.23360336)\\   =3700.81003

    On approximating,

    A=$3700.81

    (v) daily

    n=365 time per year

    A=3000[1+\frac{0.03}{365}]^{365(7)}  \\   =3000[1.000082192]^{2555}\\   =3701.002234

    On approximating the values,

    A=$3701.00

    (vi) Continuously

    A=Pe^r^t\\   =3000e^{\frac{0.03}{1}(7) }\\   =3000e^{0.21} \\   =3000(1.23367806)\\   =3701.03418\\

    On approximating the value,

    A=$3701.03

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