Answer:
a. [tex]y=6(1.7472)^x[/tex]
b. [tex]y=6e^{0.558t}[/tex]
c.13.3 months
Step-by-step explanation:
a.-Given the first term at [tex]t_0[/tex] is 6 and the second term at [tex]t_3[/tex] is 32.
-Let's take rabbit population as a function of time to be
[tex]y=ab^x[/tex]
where y is the population at time x and a the initial population at [tex]t_0\\[/tex]
#We substitute our values to calculate the value of the constant b:
[tex]y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472[/tex]
#Replace b in the population function:
[tex]y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x[/tex]
Hence, the regression for the rabbit population as a function of time x is [tex]y=6(1.7472)^x[/tex]
b. The exponential function in terms of base [tex]e[/tex] is usually expressed as:
[tex]A=A_0e^{kt}[/tex]
Where:
[tex]A_0[/tex]-is the initial population at [tex]t_o[/tex]
[tex]A[/tex]-is the population at time t.
[tex]k-[/tex]is the exponential growth constant.
[tex]e-[/tex] the exponent
Our function in terms of base exponent is rewritten as:
[tex]y=A_0e^{kt}[/tex]
#Substitute with actual figures to solve for t:
[tex]y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580[/tex]
Hence, the regression equation in terms of base e is [tex]y=6e^{0.558t}[/tex]
c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.
-We know that [tex]y=6e^{0.558t}.[/tex]
Therefore we calculate t as(take y=10001):
[tex]y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951[/tex]
Hence, it takes approximately 13.3 months for the population to exceed 10000
