Question

2 a population of prairie dogs in arizona was estimated to be 102,500 in 2014. by 2019, the population was estimated to be 94,200. which equation best models the population of prairie dogs?\\(\bigcirc\\) \\(f(x) = 94,200(0.9833)^x\\)\\(\bigcirc\\) \\(f(x) = 102,500(0.9833)^x\\)\\(\bigcirc\\) \\(f(x) = 102,500(1.017)^x\\)\\(\bigcirc\\) \\(f(x) = 94,200(1.017)^x\\)

Explanation:

Step1: Identify initial population

Initial population $P_0 = 102500$ (2014 value)

Step2: Define time interval

Time $x = 2019-2014 = 5$ years

Step3: Set up decay formula

Use $f(x)=P_0(r)^x$, substitute $f(5)=94200$:
$$94200 = 102500(r)^5$$

Step4: Solve for growth/decay factor

Rearrange to solve for $r$:
$$r^5 = \frac{94200}{102500} \approx 0.919$$
$$r = \sqrt[5]{0.919} \approx 0.983$$

Step5: Write final model

Substitute $P_0$ and $r$ into formula:
$$f(x)=102500(0.983)^x$$

Answer:

$\boldsymbol{f(x)=102,500(0.983)^x}$