Question

2.

Answer 1

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)

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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.

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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.

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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