Respuesta :
Answer:
The correct answer is p1 = 898 and p2 = 1020.
Explanation:
This is a logistic growth problem and will utilize the following formula:
[tex]Pn = P_{n-1} + r (1- \frac{P_{n-1} }{k} ) P_{n-1}[/tex]
Given ;
r = 1.7
k = 1500
Po = 400
then,
[tex]Pn = 400 + 1.7 (1- \frac{400 }{1500} ) 400[/tex]
= 400 + 1.7 (293)
= 400+ 498
P₁= 898
Similarly,
[tex]P₂= P_{2-1} + r (1- \frac{P_{2-1} }{k} ) P_{2-1}[/tex]
= [tex]Pn = 400 + 1.7 (1- \frac{898 }{1500} ) 400[/tex]
= 1020
Thus, the correct answer is p1 = 898 and p2 = 1020.
The population of fishes in the first generation will be 898 and in the second generation the population will be 1020.
Carrying Capacity:
It is the maximum number of individuals in a system at a certain time.
The logistic growth formula,
[tex]\bold {P_n = P_o + r (1- \dfrac {P_o}{k}) P_o }[/tex]
Where,
r - rate of population increase = 1.7
k- carrying capacity of the pond = 1500
Po- initial population = 400
Put the values in the formula,
[tex]\bold {P_n =400+ 1.7 (1- \dfrac {400}{1500}) 400 }\\\bold {P_n = 898 }[/tex]
In second generation,
Po = 898,
So,
[tex]\bold {P_n' = 898+ 1.7 (1- \dfrac {898}{1500}) 898}\\\\\bold {P_n' = 1020 }[/tex]
Therefore, the population of fishes in the first generation will be 898 and in the second generation the population will be 1020.
To know more about Carrying Capacity,
https://brainly.com/question/2375972
