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horsena [70]
3 years ago
6

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?
Mathematics
1 answer:
bulgar [2K]3 years ago
8 0

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

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liubo4ka [24]
He has a 1/6 chance.
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3 years ago
Complete the following statement: f(3)=_____
victus00 [196]
Using this equation, f(3) = 23.

In order to find the value of f(3), we need to take the f(x) equation and put 3 everywhere we see x. Then we follow the order of operations to solve. So, let's start with the original. 

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4 0
3 years ago
Larry has read 31 pages of his library book. The book contains p total pages.
OLEGan [10]

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Step-by-step explanation:

6 0
2 years ago
142 students are going on a field trip. There will be six drivers, and two different types of vehicles. A bus can hold 51 people
otez555 [7]

Answer: 2 buses and 4 vans would be needed

Step-by-step explanation:

Let x represent the number of buses that would be needed.

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x + y = 6

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