Respuesta :
Answer:
(A): As year increases the number of pikas reduces.
(B): As year increases the number of pikas increases as opposed to when the rate reduces.
Step-by-step explanation:
See comment for complete question
Given
[tex]a = 144[/tex] --- Initial Population
[tex]r = 8\%[/tex] --- rate
(A) WHEN THE RATE DECREASES
First, we need to write out the function when the population decreases.
This is given as:
[tex]f(x) = a(1-r)^x[/tex]
Substitute values for a and r
[tex]f(x) = 144(1-8\%)^x[/tex]
Convert % to decimal
[tex]f(x) = 144(1-0.08)^x[/tex]
[tex]f(x) = 144(0.92)^x[/tex]
Next, we calculate the average rate of change for both intervals using:
[tex]Rate = \frac{f(b) - f(a)}{b-a}[/tex]
For 1 to 5:
[tex]Rate = \frac{f(5) - f(1)}{5-1}[/tex]
[tex]Rate = \frac{f(5) - f(1)}{4}[/tex]
Calculate f(5) and f(1)
[tex]f(x) = 144(0.92)^x[/tex]
[tex]f(1) = 144*0.92^1 =144*0.92=132.48[/tex]
[tex]f(5) = 144*0.92^5 =144*0.66=95.04[/tex]
[tex]Rate = \frac{95.04 - 132.48 }{4}[/tex]
[tex]Rate = \frac{-37.44}{4}[/tex]
[tex]Rate = -9.36[/tex]
For 6 to 10:
[tex]Rate = \frac{f(10) - f(6)}{10-6}[/tex]
[tex]Rate = \frac{f(10) - f(6)}{4}[/tex]
Calculate f(6) and f(10)
[tex]f(x) = 144(0.92)^x[/tex]
[tex]f(6) = 144*0.92^6 =144*0.61=87.84[/tex]
[tex]f(10) = 144*0.92^{10} =144*0.43=61.92[/tex]
[tex]Rate = \frac{61.92-87.84}{4}[/tex]
[tex]Rate = \frac{-25.92}{4}[/tex]
[tex]Rate = -6.48[/tex]
So, we have:
[tex]Rate = -9.36[/tex] for year 1 to 5
This means that the number of pikas reduces by 9.36 yearly
[tex]Rate = -6.48[/tex] for year 6 to 10
This means that the number of pikas reduces by 6.48 yearly
So, we can say that, as year increases the number of pikas reduces.
(B) WHEN THE RATE INCREASES
First, we need to write out the function when the population decreases.
This is given as:
[tex]f(x) = a(1-r)^x[/tex]
Substitute values for a and r
[tex]f(x) = 144(1+8\%)^x[/tex]
Convert % to decimal
[tex]f(x) = 144(1+0.08)^x[/tex]
[tex]f(x) = 144(1.08)^x[/tex]
Next, we calculate the average rate of change for both intervals using:
[tex]Rate = \frac{f(b) - f(a)}{b-a}[/tex]
For 1 to 5:
[tex]Rate = \frac{f(5) - f(1)}{5-1}[/tex]
[tex]Rate = \frac{f(5) - f(1)}{4}[/tex]
Calculate f(5) and f(1)
[tex]f(x) = 144(1.08)^x[/tex]
[tex]f(1) = 144(1.08)^1 = 144*1.08= 155.52[/tex]
[tex]f(5) = 144(1.08)^5 = 144*1.47= 211.68[/tex]
[tex]Rate = \frac{211.68 - 155.52}{4}[/tex]
[tex]Rate = \frac{56.16}{4}[/tex]
[tex]Rate = 14.04[/tex]
For 6 to 10:
[tex]Rate = \frac{f(10) - f(6)}{10-6}[/tex]
[tex]Rate = \frac{f(10) - f(6)}{4}[/tex]
Calculate f(6) and f(10)
[tex]f(x) = 144(1.08)^x[/tex]
[tex]f(6) = 144(1.08)^6 = 228.52[/tex]
[tex]f(10) = 144(1.08)^{10} = 310.89[/tex]
[tex]Rate = \frac{310.89-228.52}{4}[/tex]
[tex]Rate = \frac{82.37}{4}[/tex]
[tex]Rate = 20.59[/tex]
So, we have:
[tex]Rate = 14.04[/tex] for year 1 to 5
This means that the number of pikas increases by 14.04 yearly
[tex]Rate = 20.59[/tex] for year 6 to 10
This means that the number of pikas increases by 20.59 yearly
So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.
