Question

The population of Whoville has been decreasing at a rate of 0.8% per year since Dr. Seuss passed away in 1991. If the population was 13,500 at the beginning of 2005, which expression gives its population at the end of 1998?

Answer 1

Answer:

Using P(7) = 13500 in the expression $P(t) = P(0)(0.992)^{t}$, we find that the population at the end of 1998 is given by the expression $P(0) = \frac{13500}{(0.992)^{7}}$ 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.

$P(t) = P(0)(1 - r)^{t}$

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 $r = 0.008$

Then

$P(t) = P(0)(1 - 0.008)^{t}$

$P(t) = P(0)(0.992)^{t}$

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).

$P(t) = P(0)(0.992)^{t}$

$13500 = P(0)(0.992)^{7}$

$P(0) = \frac{13500}{(0.992)^{7}}$

$P(0) = 14281$

Using P(7) = 13500 in the expression $P(t) = P(0)(0.992)^{t}$, we find that the population at the end of 1998 is given by the expression $P(0) = \frac{13500}{(0.992)^{7}}$ and it was of 14,281.

Answer 2

Well, if your asking from 2005 back to 1998... Starting from 13,500 take back 0.8% each year, in this case 7 years, it would be about 2831.