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 11 views 0

Answers ( )

  1. ThanhThu
    0
    2021-07-30T23:20:58+00:00
    Reply

    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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