Given the function:
[tex]P(t)=510*0.91^t[/tex]To solve this question, follow the steps below.
Step 01: Find the initial population size.
The initial population size is the population size when t = 0.
Then, substitute x by 0 to find P(0).
[tex]\begin{gathered} P(0)=510*0.91^0 \\ P(0)=510*1 \\ P(0)=510 \end{gathered}[/tex]The initial population size is 510 fish.
Step 02: Find it the population size is increasing or decreasing.
To do it, find P(1) and P(2):
[tex]\begin{gathered} P(1)=510*0.91^1 \\ P(1)=510*0.91 \\ P(1)=464.1 \end{gathered}[/tex][tex]\begin{gathered} P(2)=510*0.91^2 \\ P(2)=510*0.828 \\ P(2)=422.3 \end{gathered}[/tex]So, P(0) < P(1) < P(2). The population is decreasing.
The function represents a decay.
Step 03: Find the population change per year.
To find the population change, use the formula below:
[tex]C(\%)=\frac{P(t+1)-P(t)}{P(t)}*100[/tex]So, comparing P(0) and (P1):
[tex]\begin{gathered} C=\frac{464.1-510}{510}*100 \\ C=-0.09*100 \\ C=-9\% \end{gathered}[/tex]The population decreases by a percent of 9%.
In summary:
- The initial population size is 510 fish.
- The function represents a decay.
- The population decreases by a percent of 9%.
