Question

A population of fish in a lake is 14000 in 2010. The population decreases 6% annually. What is the population in 2019

Answer 1

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: $y=a*b^x$, 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.

$6\5=\frac{6}{100}=0.06$

Upon substituting our given values in exponential form of function we will get the population of fish in the lake after x years as:

$y=14,000*(1-0.06)^x$

$y=14,000*(0.94)^x$

Let us find x by subtracting 2010 from 2019.

$x=2019-2010=9$

Upon substituting x=9 in our function we will get,

$y=14,000*(0.94)^9$

$y=14,000*0.5729948022286167$

$y=8021.927\approx 8022$

Therefore, the population of fish in 2019 will be 8022.