Respuesta :
Let A represent the value of the car after each year.
A= initial value (P)×(1+percent increase(r)) ^time
A=P×(1+r)^t
A=18710×(1+(-12%))^8
A=18710×(1-12%)^8
A=18710×(1-0.12)^8
A=18710×(0.88)^8
A= 6728.7619591115
The best approximation is 6729
Therefore the value of the car will be about $6729 after 8 years
Your answer is B.
A= initial value (P)×(1+percent increase(r)) ^time
A=P×(1+r)^t
A=18710×(1+(-12%))^8
A=18710×(1-12%)^8
A=18710×(1-0.12)^8
A=18710×(0.88)^8
A= 6728.7619591115
The best approximation is 6729
Therefore the value of the car will be about $6729 after 8 years
Your answer is B.
The value of her car at the end of 8 years is approximately $6,256.9
Data;
- Principal = $18,710
- annual depreciation = 12%
- number of years = 8 years.
Straight line Depreciation
To solve this problem, we have to understand that this car value is calculated using straight line depreciation method. This means that it's value decreases by a constant value at the end of each year
for year 1
12% of 18,710
[tex]0.12 * 18710 = 2245.2\\18710 - 2245.2 = 16464.8[/tex]
for year 2
12% of 16464.8
[tex]0.12 * 16464.8 = 1975.776\\16464.8 - 1975.776 = 14489.024[/tex]
for year 3
12% of 14489.024
[tex]0.12* 14489.024= 1738.682\\14489.024 - 1738.862 = 12750.34[/tex]
for year 4
12% of 12750.34
[tex]0.12 * 12750.34 = 1530.040\\12750.34 -1530.040 = 10920.299\\[/tex]
for year 5
12% of 10,920.299
[tex]0.12 * 10920.299 = 1310.435\\10,920.299 - 1310.435 = 9609.86[/tex]
for year 6
12% of 9609.86
[tex]0.12 * 9609.86 = 1153.18\\9609.86 - 1153.18 = 8079.68\\[/tex]
for year 7
12% of 8079.68
[tex]0.12 * 8079.68 = 969.562\\8079.68 - 969.562 = 7110.118[/tex]
for year 8
12% of 7110.118
[tex]0.12 * 7110.118 = 853.214\\7110.118 - 853.214 = 6256.904[/tex]
The value of her car at the end of 8 years is approximately $6,256.9
Learn more on straight line depreciation method here;
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