Answer:
6000
Step-by-step explanation:
If the fish population in a local lake increases at a rate proportional to the population each moment, this can be expressed as;
P = k/t (the scenario is an inverse proportionality because as the time is reducing, the population is increasing by the same factor)
If there are 1000 fish 5 years ago and there are 2000 fish 4 years ago
Yearly percentage increase will be;
2000-1000/1000 × 100%
1000/1000×100%
= 100%
This means that the population increases by 100% on a yearly basis (by a factor of 1000)
The population 3 years ago will be population
P = 2000+1000
P = 3000
The population 2 years ago will be population
P = 3000+1000
P = 4000
The population 1 year ago will be population
P = 4000+1000
P = 5000
The population now will be population
P = 5000+1000
P = 6000 populations
