Question

A population of a certain species of snake is 28,000 animals and is decreasing by 200 snakes annually.Write an equation for y, the population at time t (in years), representing the situation.y =How many snakes are in the population after 20 years?

Answer 1

Step-by-step 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;

`a_n=a_1+(n-1)d`

Take note of the variables here as;

`\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}`

To determine the snake population at any given year, we would come up with the following formula;

`a_n=28000+(n-1)(-200)`
`a_n=28000+(-200n+200)`

Therefore the equation representing the situation using y as the population and t as the time in years will be;

`y=28000+(-200t+200)`

To now determine how many snakes are in the population after 20 years, we now substitute 20 in place of t;

`\begin{gathered} Population\text{ }after\text{ }20\text{ }years: \\ y=28000+(-200(20)+200) \end{gathered}`
`y=28000+(-4000+200)`
`y=28000+(-3800)`
`y=24200`

ANSWER:

`\begin{gathered} Equation: \\ y=28000+(-200t+200) \\ Population\text{ }after\text{ }20\text{ }years: \\ y=24200 \end{gathered}`