The table below shows the estimated population of fish living in the local lake as determined by the local environment council every 10 years between 1960 and 1990.

Question

The table below shows the estimated population of fish living in the local lake as determined by the local environment council every 10 years between 1960 and 1990.

The equation y=1500(.9)x describes the curve of best fit for the fish population (y). Let x represent the number of 10–year intervals since 1960.

Year Population

1960 1500

1970 1350

1980 1210

1990 1100

Using this equation, what is the approximate predicted population of fish living in the lake in the year 2020?

a) 800

b) 980

c)720

d) 770

I chose the answer a) because eat year the fish are decreasing by .9 percent and the answer I got for how many fish there would be was: 801.9 which is pretty close to 800 and this approximate answer.

Answer 1

Oh my...

0.9 is 90% aaaand we're looking at every 10 years. If the population is 90% of what it was 10 years ago, that's a 10% decrease.

The way to solve your problem is by realizing that 2020 - 1960 = 60, so that's 6 decades.

y = 1500(0.9)⁶ = 797 ≈ 800

So, yes, your answer is a.