The fish population at the beginning of 2012 is 100.
A fish population, p, can be represented by the equation;
[tex]\rm P =800 \left (\dfrac{1}{2} \right)^t[/tex]
Where t is time in years since the beginning of 2015.
The expression represents the number of population in the year 2015 so first take the difference between the year 2015 and 2012 which is 3 then substitute the 3 in the given equation and solve.
The difference between year 2015 and 2012 is;
t = 2015 - 2012 = 3
Substitute the value of t in the equation
[tex]\rm P =800 \left (\dfrac{1}{2} \right)^t\\\\\rm P =800 \left (\dfrac{1}{2} \right)^3\\\\P = 800 \times \dfrac{1}{8}\\\\P = 100[/tex]
Hence, the fish population at the beginning of 2012 is 100.
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