Answer:
In 1981, the Australian humpback whale population was 350
Po = Initial population = 350
rate of increase = 14% annually
P(t) = Po*(1.14)^t
P(t) = 350*(1.14)^t
Where
t = number of years that have passed since 1981
Year 2000
2000 - 1981 = 19 years
P(19) = 350*(1.14)^19
P(19) = 350*12.055
P(19) = 4219.49
P(19) ≈ 4219
Year 2018
2018 - 1981 = 37 years
P(37) = 350*(1.14)^37
P(37) = 350*127.4909
P(37) = 44621.84
P(37) ≈ 44622
There would be about 44622 humpback whales in the year 2018
67
Step-by-step explanation:
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x=3×3
ok it's fine if you don't mind
the answer to your question is 'b'
2.14 X 10(power4)
D.
-2*-3=6 -2*5=-10. -2*1=-2
-2*8=-16. -2*-2=4. -2*3=-6
-2*2=-4. -2*1=-2. -2*-4=8