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
Answer:
Step-by-step explanation:
The number of defective laptops = 6/145 * 100 as a percent.
The number of defective laptops = 4.13% in a sample of 145
A: False. The percentage should be the same for 290 laptops. The % does not change. The number of defective laptops will change (12)
B: True. It is a reason assertion that about 4% of all laptops will be defective.
C: True. It is not reasonable to assume that only 0.5 laptops will be defective. .5% is too small.
D: True. 10% is too high.
E: False. The variation should not be much.
1/4
if 3/4 cyclists finish then that leaves 1/4 of the group that did not finish
The second option!! the addition one :))
