Question

the function shown in the graph represents the number of lions in a region after x years, where the rate of decay is 20%. the number of zebras in that same region after x years can be modeled by the function ( f(x)=3000(0.8)^{x} ). a representative for a conservationist group claims there will be fewer lions than zebras within 2 years. is the representative correct? justify your answer. the representative is incorrect. after 2 years, the exponential model for the zebra population predicts there will be approximately (square) zebra(s) in the region and an exponential model for the lion population predicts there will be approximately (square) lion(s) in the region (round to the nearest whole number as needed.)

Explanation:

Step1: Substitute \(x = 2\) into the function \(f(x)=3000(0.8)^{x}\)

The function for the population of zebras is \(f(x)=3000(0.8)^{x}\), where \(x\) is the number of years. When \(x = 2\), we use the formula \(a\times b^{x}\). Here \(a = 3000\), \(b=0.8\) and \(x = 2\).

$$f(2)=3000\times(0.8)^{2}$$

Step2: Calculate \((0.8)^{2}\)

Using the exponent rule \(b^{n}=\underbrace{b\times b\times\cdots\times b}_{n\text{ times}}\), for \(b = 0.8\) and \(n = 2\), we have \((0.8)^{2}=0.8\times0.8 = 0.64\)

$$f(2)=3000\times0.64$$

Step3: Calculate the final product

$$3000\times0.64=(3\times10^{3})\times(6.4\times10^{- 1})=(3\times6.4)\times10^{3 - 1}=19.2\times10^{2}=1920$$

Answer:

After \(2\) years, the exponential model for the zebra population predicts there will be approximately \(1920\) zebra(s) in the region.