Suppose the fish population in a local lake increases at a rate proportional to the population each moment. there were 1000 fish 5 years ago and 2000 fish 4 years ago. How many fish are there right now?

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

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6