Respuesta :
EXPLANATION:
Given;
We are told that the initial population of a certain snake species is 28,000. Also the snakes are decreasing at the rate of 200 snakes annually.
Required;
We are required to write an equation showing the population at time t.
Step-by-step solution;
Note that what we have here is an arithmetic progression. The difference in population between one year and the next is a fixed figure and that is 200.
The general formula for an arithmetic progression is;
[tex]a_n=a_1+(n-1)d[/tex]Take note of the variables here as;
[tex]\begin{gathered} a_1=initial\text{ }population\text{ }(first\text{ }term) \\ d=annual\text{ }decrease\text{ }(common\text{ }difference) \\ n=the\text{ }nth\text{ }term \end{gathered}[/tex]To determine the snake population at any given year, we would come up with the following formula;
[tex]a_n=28000+(n-1)(-200)[/tex][tex]a_n=28000+(-200n+200)[/tex]Therefore the equation representing the situation using y as the population and t as the time in years will be;
[tex]y=28000+(-200t+200)[/tex]To now determine how many snakes are in the population after 20 years, we now substitute 20 in place of t;
[tex]\begin{gathered} Population\text{ }after\text{ }20\text{ }years: \\ y=28000+(-200(20)+200) \end{gathered}[/tex][tex]y=28000+(-4000+200)[/tex][tex]y=28000+(-3800)[/tex][tex]y=24200[/tex]ANSWER:
[tex]\begin{gathered} Equation: \\ y=28000+(-200t+200) \\ Population\text{ }after\text{ }20\text{ }years: \\ y=24200 \end{gathered}[/tex]