Answer: 1.C) 10% (approx)
2. B) Exponential
Step-by-step explanation:
1) Since, The initial price of the car, P = $24,000
And, the total time, t = 6.8 years.
Final price of the car after 6.8 years, A = 12,000
Let the rate of depreciation is r %
Thus, [tex]A= P (1-\frac{r}{100})^{6.58}[/tex]
⇒ [tex]12,000= 24,000 (1-\frac{r}{100})^{6.58}[/tex]
⇒ [tex]0.5 = (1+\frac{r}{100})^{6.58}[/tex]
⇒ [tex](0.5)^{1/6.58} = 1-\frac{r}{100}[/tex]
⇒ [tex](0.5)^{1/6.58} = 1-\frac{r}{100}[/tex]
⇒[tex]0.90001709913 = 1-\frac{r}{100}[/tex]
⇒ [tex]r=100-90.001709913=9.99829008677\approx 10\%[/tex]
Thus, Option C is correct.
2) Here, The price of car initially, P = $20,000
Rate of decay, r = 20%
Thus, the function that shows the price of car,
[tex]f(x)= 20,000 (1-\frac{20}{100})^x[/tex]
[tex]f(x)= 20,000 (1-\frac{1}{5})^x[/tex]
[tex]f(x)= 20,000 (1.2)^x[/tex]
Since, the function is decreasing by the constant percent not the constant rate.
Therefore,By the property of exponential function,
f(x) is an exponential function.
Thus, Option B is correct.
