If the population of ants is 500 today, which function models the population at x months?

f (x) = 500 • 1.1x

f (x) = 1.1 • 500x

f (x) = 500 • 0.9x

f (x) = 0.9 • 500x

f (x) = 500 • 1.1x

Step-by-step explanation:

To solve the question, we need to understand that:

- The population of ants is going to increase with time as they mate and have a population positive growth rate.

- The initial number of ants is 500.

- The rate at which it increases will be multiplied with x months

We will work our way out by removing the wrong answers first

Since,

The population cannot be decreased so we can neglect the last two options involving 0.9 as it will result in a decrease in the population.

Now we are left with two options:

a) f (x) = 500 • 1.1x

b) f (x) = 1.1 • 500x

Since in the option (b), 500 is multiplied with x, it is not possible as 500 is a constant initial amount and not a factor of growth of population. The growth factor is supposed to be multiple of x months.

So, the correct answer is f (x) = 500 • 1.1x