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Sagot :
Answer:
This situation represents [a.) exponential decay
The rate of growth or decay, r, is equal to c.) 0.2
So the value of the computer each year is g.) 80% of the value in the previous year.
It will take [i.) 3 years for the value of the computer to reach $512.
Step-by-step explanation:
Since the value depreciates each year, it represents exponential decay.
The rate of growth or decay, r, is equal to
Depreciates 20%, so r = 0.2.
So the value of the computer each year is ... of the value in the previous year,
Depreciates 20%, so it is 100% - 20% = 80% of the value in the previous year.
It will take x years for the value of the computer to reach $512.
The value of the computer after x years is given by:
[tex]V(x) = V(0)(1-r)^x[/tex]
In which V(0) is the initial value and r is the decay rate. So
[tex]V(x) = 1000(1-0.2)^x[/tex]
[tex]V(x) = 1000(0.8)^x[/tex]
We want to find x for which V(x) = 512. So
[tex]V(x) = 1000(0.8)^x[/tex]
[tex]512 = 1000(0.8)^x[/tex]
[tex](0.8)^x = \frac{512}{1000}[/tex]
[tex](0.8)^x = (0.512)[/tex]
Applying log to both sides:
[tex]\log{(0.8)^x} = \log{(0.512)}[/tex]
[tex]x\log{0.8} = \log{(0.512)}[/tex]
[tex]x = \frac{\log{0.512}}{\log{0.8}}[/tex]
[tex]x = 3[/tex]
So it will take 3 years.
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