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Answers:
a) 0.0625
b) 0.9375
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Work Shown:
The probability of landing on heads is 1/2 = 0.5 since both sides are equally likely to land on. Getting 4 heads in a row is (1/2)^4 = (0.5)^4 = 0.0625
The event of getting at least one tail is the complement of getting all four heads. This is because you either get all four heads or you get at least one tail. One or the other must happen. We subtract the result we got from 1 to get 1-0.0625 = 0.9375
You can think of it like this
P(getting all four heads) + P(getting at least one tail) = 1
The phrasing "at least one tail" means "one tail or more".
Answer:
36 and 49
Step-by-step explanation:
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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
