Answer :
Answer:
8022.
Step-by-step explanation:
Let x be the number of years after 2010.
We have been given a population of fish in a lake is 14000 in 2010. The population decreases 6% annually.
We can see that population of fish is the lake is decreasing exponentially as it is decreasing 6% annually.
Since we know that an exponential function is in form: [tex]y=a*b^x[/tex], where,
a = Initial value,
b = For decrease or decay b is in form (1-r) where r represents decay rate in decimal form.
Let us convert our given decay rate in decimal form.
[tex]6\5=\frac{6}{100}=0.06[/tex]
Upon substituting our given values in exponential form of function we will get the population of fish in the lake after x years as:
[tex]y=14,000*(1-0.06)^x[/tex]
[tex]y=14,000*(0.94)^x[/tex]
Let us find x by subtracting 2010 from 2019.
[tex]x=2019-2010=9[/tex]
Upon substituting x=9 in our function we will get,
[tex]y=14,000*(0.94)^9[/tex]
[tex]y=14,000*0.5729948022286167[/tex]
[tex]y=8021.927\approx 8022[/tex]
Therefore, the population of fish in 2019 will be 8022.