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:
-6
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
4(2x+y)
8x + 2y
Answer:
6 
x 8
Step-by-step explanation:
We'll represent the number with the expression 4x + 7
Then, we'll set up the inequality:
31 4x + 7 55
24 4x 48
6 x 8
Answer:
C = 65
Step-by-step explanation:
A + B = 180
(2x) + (3x - 20) = 180
5x - 20 = 180
5x = 200
x = 40
Substitute for 32 in for x:
2x - 15 = C
2(40) - 15 = C
80 - 15 = C
65 = C
