Answer:
1. Small expenditures which primarily benefit the current period. REVENUE EXPENDITURES
2. Cost less accumulated depreciation. BOOK VALUE
3. An accelerated depreciation method used for financial statement purposes. DOUBLE DECLINING BALANCE METHOD
4. Tangible resources that are used in operations and are not intended for resale. PLANT ASSETS
5. Equal amount of depreciation each period. STRAIGHT LINE METHOD
6. Expected cash value of the asset at the end of its useful life. SALVAGE VALUE
7. Process of allocating the cost of equipment over its service life. DEPRECIATION
8. Material expenditures that increase an asset's operating efficiency, productive capacity, or useful life CAPITAL EXPENDITURES
9. An accelerated depreciation method used for tax purposes. MACRS
10. Useful life is expressed in terms of units of production or expected use. UNITS OF ACTIVITY METHOD
Explanation:
Answer:
number of periods = 8 years.
Explanation:
We know,
Future Value = Present value ×
Here,
Present value = PV = $2,500
Future value = FV = $3,500
Interest rate (Compounding) = 5% = 0.05
We have to determine how many years (Periods) it will take, n = ?
Putting the values into the above formula,
$3,500 = $2,500 ×
or, = $3,500 ÷ $2,500
or, n log 1.05 = 1.4
or, n × 0.17609 = 1.4
or, n = 1.4 ÷ 0.17609
Therefore, number of years = 7.95 or 8 years.
Answer:
a. $965.74
b. $939.11
Explanation:
In this question we use the Present value formula i.e shown in the attachment below:
1. Given that,
Future value = $1,000
Rate of interest = 6.5%
NPER = 4 years
PMT = $1,000 × 5.5% = $55
The formula is shown below:
= -PV(Rate;NPER;PMT;FV;type)
So, after solving this, the price would be $965.74
2. Given that,
Future value = $1,000
Rate of interest = 6.5%
NPER = 8 years
PMT = $1,000 × 5.5% = $55
The formula is shown below:
= -PV(Rate;NPER;PMT;FV;type)
So, after solving this, the price would be $939.11
Answer:
V(n)=140,000-10000n
V(7)=$70,000
Explanation:
Purchase Cost= $140,000
Value After 11 Years =$30,000
Depreciation per Year =
The truck depreciates at a rate of $10000 per year.
Using straight-line depreciation, the value of the truck in dollars, V
The linear function of its age in years n, V(n)=140,000-10000n
When the truck is 7 years old
n=7
Truck's Value, V(n)=140,000-10000n
=140,000-(10000X7)
=140,000-70000
=$70,000