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:
B. Yes, because 0.004<0.010.004<0.010, point, 004, is less than, 0, point, 01.
Step-by-step explanation:
Khan Academy
Answer:
12
Step-by-step explanation:
a2 b
( )2 ( )
(–2)2 (3)
(4)(3)
So the ration is 1 to 5 since it's one foot every 5 seconds. So for every 5 seconds, add 1 foot.
For example: in 45 seconds, it would travel 7 feet.
But in this Scenario, the timer starts after 5 feet so just add 5
Ex. in 45 seconds, it would have traveled 12 feet since we added the 5 feet that we started with to the 7 other feet it traveled.
Idk If this will help very much but I hope it does.
You multiply 16 by 11 to get k, which is 176!
