Answer: Options 2, 4 and 5
Step-by-step explanation:
E2020 Answers
7^5. Found by trial and error.
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:
30000 times larger.
Step-by-step explanation:
10^7 * 9/10^3*3
Law of exponents cancel some of the tens.
10^4 * 9/3
Cancel out the three.
10^4 *3
Compute
30000 times larger.
That would be C because there are two zeros right?
So you move two decimal places from 67.
And you get 0.67
