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
Answer:
4⁻⁴
Step-by-step explanation:
4⁵ × 4⁻⁷ ÷ 4⁻²
⁺ ⁵ ⁻ ⁷
= ⁻²
⁻ ² ⁻ ²
= ⁻ ⁴
4⁻⁴
Answer:6
4 multiplied by 6 is 24 so 6 is the answer
Answer:
[-2, 4]
Step-by-step explanation:
The domain of a function is the set of all x-values a relation could possibly take on to have an output. This is represented by in the graph by having the line correspond to the x-value. From the graph shown we can see that x-values from -2 to 4 have a line corresponding to them, and the dots are closed meaning it is inclusive. This would be written as: [-2, 4]
