Answer: 0.70
Step-by-step explanation:
Given : The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:
x 0 1 2 3 4 5
P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05
Using the above probability distribution , the the probability that in a given week there will be at most 3 accidents is given by :_
Hence, the required probability = 0.70
128 : 3 :: w : 8 proportion
3w = (8)(128) product means/extremes
w = 341 ⅓ cars
Answer:
Step-by-step explanation:
Given that,
y' = 17y ( 1-y^7)
Let y=1
Then, y' = 0 for all t
Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.
So, the graph solution will be
y(0) = 1 and this will be an horizontal line
If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.
Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.
So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.
This make the limit be 1.
This shows that x = 1 is a stable equilibrium.