Step-by-step explanation:
(a) dP/dt = kP (1 − P/L)
L is the carrying capacity (20 billion = 20,000 million).
Since P₀ is small compared to L, we can approximate the initial rate as:
(dP/dt)₀ ≈ kP₀
Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.
20 = k (6,100)
k = 1/305
dP/dt = 1/305 P (1 − (P/20,000))
(b) P(t) = 20,000 / (1 + Ce^(-t/305))
6,100 = 20,000 / (1 + C)
C = 2.279
P(t) = 20,000 / (1 + 2.279e^(-t/305))
P(10) = 20,000 / (1 + 2.279e^(-10/305))
P(10) = 6240 million
P(10) = 6.24 billion
This is less than the actual population of 6.9 billion.
(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))
P(100) = 7570 million = 7.57 billion
P(600) = 20,000 / (1 + 2.279e^(-600/305))
P(600) = 15170 million = 15.17 billion
5000 rounded to the nearest hundredth is 5000
The answer should be 2 13/30.
11, count everything after 3 and everything before 4
Answer:
P(x)=-2x^2+34x+98
The cost function is given by c(x)=21x-98
And the revenue function is given by r(x)=55x-2x^2
The profit function is the difference of revenue function and the cost function. Mathematically, we can write
p(x)=r(x)-c(x)
Substituting the values, we get
p(x)=55x-2x^2-(21x-98)
p(x)=-2x^2 + 34x = 98
Therefore, the profit function is
p(x) = -2x^2 + 34x + 98
Step-by-step explanation:
