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
Yes . Mark me brainliest !!??
Answer:
x=-15/2
Step-by-step explanation:
Answer:
So 3 for $1.26 and 4 for $1.68
Step-by-step explanation:
3 for $1.26 means $0.42 per each
4 for $1.68 means $0.42 per each
5 for $2.25 means $0.45 per each
So 3 for $1.26 and 4 for $1.68 would be the least expensive unit rate!
Hope this help :3
