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Nina [5.8K]
3 years ago
9

A piece of equipment is purchased for $18,000.00 and has a salvage value of $2,000.00. The estimated life is 10 years and the me

thod of depreciation is straight-line. Shipping costs total $620.00 and installation costs are $590.00. The depreciation for year 2 is
$1721.00
$1800.00
$200.00
$1921.00
Mathematics
1 answer:
const2013 [10]3 years ago
6 0
The full $1921.00 because of company incrued cost
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  • $1,094,748.09
  • $8600.28
  • 11 years, 10 months

Step-by-step explanation:

a) The compound interest formula can be used:

  A = P(1 +r/n)^(nt)

where P is the principal invested at annual rate r compounded n times per year for t years.

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__

b1) The amortization formula is good for this.

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the amount earning interest at annual rate r compounded n times per year for t years.

  A = $1,094,748.09(0.049/12)/(1 - (1 +0.049/12)^(-12·15)) ≈ $8600.28

Tamsyn can withdraw $8600.28 per month for 15 years.

__

b2) The account balance after n months will be ...

  B = P(1 +r/12)^n -A((1+r/12)^n -1)/(r/12)

Filling in the known values and solving for n, we have ...

  300,000 = 1,094,748.09(1.1.00408333^n) -8600.28(1.00408333^n -1)/.000408333

  300,000 = 1,094,748.09(1.1.00408333^n) -2,106,191.02(1.00408333^n -1)

  -1,806,191.02 = -1,011,442.93(1.00408333^n)

  1.785757 = 1.00408333^n

  n = log(1.785757)/log(1.00408333) = 142.3

After about 11 years 10 months, the account balance will be $300,000.

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