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
The answer is 62^ it reflects angle IK
Answer:
Sam is incorrect.
Step-by-step explanation:
Using Pythagorean Theorem, we can find the length of diagonal SQ as 13.4 (6^2 + 12^2 = c^2, 36 + 144 = c^2, sqrt(180) = c, c is approx 13.4). We can do the same for diagonal OM (6^2 + 6^2 = c^2, 36 + 36 = c^2, 72 = c^2, sqrt(72) = c, c is approx 8.5). Sam is therefore incorrect because 13.4 is not double of 8.5.
I think the answer would be C
As it is basically a straight line so there wouldn't be necessary for there to be a scatter diagram for it.
I may be wrong so take it with a grain of salt
Answer:
8
Step-by-step explanation:
