Answer:
we conclude that the intervals in which the function P(t) is decreasing are:
Step-by-step explanation:
At x< 1, the function P(t) is decreasing
- A function p is an increasing function on an open interval if f(y) > f(x) for any two input values x and y in the given interval where y>x
- A function p is a decreasing function on an open interval if f(y) > f(x) for any two input values x and y in the given interval where y>x
From the figure, it is clear that the function seems to be increasing
from (1, 3) and then 4 to onwards.
But it is clear that the function seems to be decreasing from the x < 1 and from the interval (3, 4).
Therefore, we conclude that the intervals in which the function P(t) is decreasing are:
D is true hope this helps
Answer:
I think 4
Step-by-step explanation:
A variable term is a term with a variable on it or a letter.
3x, 3y, 5x, and y are all variable terms
Part (a)
P(A) = 0.5
P(B) = 0.4
P(B/A) = 0.6
P(A and B) = P(A)*P(B/A)
P(A and B) = 0.5*0.6
P(A and B) = 0.3
<h3>Answer: 0.3</h3>
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Part (b)
We'll use the result from part (a)
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.5 + 0.4 - 0.3
P(A or B) = 0.6
<h3>Answer: 0.6</h3>
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Part (c)
A and B are not independent since P(B) does not equal P(B/A). The fact that event A happens changes the probability P(B). Recall that P(B/A) means "probability P(B) based on event A already happened". A and B are independent if P(B) = P(B/A).
Events A and B are not mutually exclusive since P(A or B) is not zero.
<h3>Answer: Neither</h3>
