Respuesta :
Answer:
[tex] D= \frac{1}{11} 15966 = 1451.454[/tex]
And we can conclude that with this method the depreciation each year is about 1451.454$ for the item analyzed.
And we can construct the following table:
year Depreciation Ramining value
0 X 15966
1 1451.454 14514.545
2 1451.454 13063.091
3 1451.454 11611.638
4 1451.454 10160.183
5 1451.454 8708.729
6 1451.454 7257.275
7 1451.454 5805.821
8 1451.454 4354.367
9 1451.454 2902.913
10 1451.454 1451.454
11 1451.454 0
Step-by-step explanation:
Previous concepts
Straight line depreciation is "the default method used to recognize the carrying amount of a fixed asset evenly over its useful life".
Solution for the problem
We assume that the useful life for this problem is 11 years
For this case we have the following function in terms of th years (n) and the initilal price (x)
[tex] D= \frac{1}{n} x[/tex]
For this case we have the following values:
[tex] x = 15966[/tex] represent the initial cost for a item
[tex] n = 11[/tex] the year on which we want to find the depreciation
And if we replace these values into our function we got this:
[tex] D= \frac{1}{11} 15966 = 1451.454[/tex]
And we can conclude that with this method the depreciation each year is about 1451.454$ for the item analyzed.
And we can construct the following table:
year Depreciation Ramining value
0 X 15966
1 1451.454 14514.545
2 1451.454 13063.091
3 1451.454 11611.638
4 1451.454 10160.183
5 1451.454 8708.729
6 1451.454 7257.275
7 1451.454 5805.821
8 1451.454 4354.367
9 1451.454 2902.913
10 1451.454 1451.454
11 1451.454 0