Replacing into the equation with t = 0, 1 and 14, we get:
[tex]\begin{gathered} P(t)=2600\cdot(\frac{96}{100})^t \\ P(0)=2600\cdot(\frac{96}{100})^0=2600\cdot1=2600 \\ P(1)=2600\cdot(\frac{96}{100})^1=2600\cdot\frac{96}{100}=2496 \\ P(14)=2600\cdot(\frac{96}{100})^{14}\approx2600\cdot0.565\approx1468 \end{gathered}[/tex]
And the table is:
t | P(t)
0 | 2600
1 | 2496
14 | 1468
As we can see in the table, the population is decreasing. This is because the base of the exponent, c = 96/100, is less than 1.
The statement P(14) means that t = 14, that is, 14 years have elapsed since 2000. And the population is 1468 students.
