Answer:
Using P(7) = 13500 in the expression [tex]P(t) = P(0)(0.992)^{t}[/tex], we find that the population at the end of 1998 is given by the expression [tex]P(0) = \frac{13500}{(0.992)^{7}}[/tex] and it was of 14,281.
Step-by-step explanation:
The population od Whoville in t years after 1998 is given by the following equation.
[tex]P(t) = P(0)(1 - r)^{t}[/tex]
In which P(0) is the population in 1998 and r is the constant rate that it decreases, as a decimal.
The population of Whoville has been decreasing at a rate of 0.8% per year since Dr. Seuss passed away in 1991.
So [tex]r = 0.008[/tex]
Then
[tex]P(t) = P(0)(1 - 0.008)^{t}[/tex]
[tex]P(t) = P(0)(0.992)^{t}[/tex]
If the population was 13,500 at the beginning of 2005, which expression gives its population at the end of 1998?
2005 is 2005-1998 = 7 years after 1998. So p(7) = 13500. We have to find P(0).
[tex]P(t) = P(0)(0.992)^{t}[/tex]
[tex]13500 = P(0)(0.992)^{7}[/tex]
[tex]P(0) = \frac{13500}{(0.992)^{7}}[/tex]
[tex]P(0) = 14281[/tex]
Using P(7) = 13500 in the expression [tex]P(t) = P(0)(0.992)^{t}[/tex], we find that the population at the end of 1998 is given by the expression [tex]P(0) = \frac{13500}{(0.992)^{7}}[/tex] and it was of 14,281.
