Turn the information into coordinate points.

Point = (time, population)

Point 1 = (1985, 45000)
Point 2 = (2004 , 26000)

Find the slope between these points using the formula

( Slope)—> m = (y2 - y1) / (x2 - x1)
26000-45000/2004-1985=-19000/19= -1000
Average rate of change is decrease of 1000 sea lion per year

I hope that helped

6 0

2 years ago

abruzzese [7]

Answer:

a) 30 kangaroos in 2030

b) decreasing 8% per year

c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo

Step-by-step explanation:

We assume your equation is supposed to be ...

P(t) = 76(0.92^t)

a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30

In the year 2030, the population of kangaroos in the province is modeled to be 30.

b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.

The population is decreasing by 8% each year.

c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.

P(100) = 75(0.92^100) = 76(0.0002392)

P(100) ≈ 0.0182, about 1/55th of a kangaroo

5 0

3 years ago