Answer:
Probability that an international flight leaving the US is delayed in departing given that the flight is a transpacific flight is 0.2105.
Step-by-step explanation:
We are given that the probability that an international flight leaving the United States is delayed in departing (event D) is 0.25. The probability that an international flight leaving the United States is a transpacific flight (event P) is 0.57. The probability that an international flight leaving the U.S. is a transpacific flight and is delayed in departing is 0.12.
Let the probability that an international flight leaving the United States is delayed in departing = P(D) = 0.25
Probability that an international flight leaving the United States is a transpacific flight = P(P) = 0.57
Probability that an international flight leaving the U.S. is a transpacific flight and is delayed in departing = = 0.12
Now, the probability that an international flight leaving the US is delayed in departing given that the flight is a transpacific flight is given by = P(D/P)
As we know that the conditional probability is given by;
P(A/B) =
Similarly, P(D/P) =
=
= 0.2105
<em>Hence, the required conditional probability is 0.2105.</em>
First find the slope of the line so change in y over change in x so 4-2/1-2 which is 2/-1 so -2 is your slope then plug that as well as a point from above to find your y intercept so y=mx+b 2=-2(2)+ b then you have 2=-4+b then add 4 to each side and you find b is six then you make ur equation from your slope and y int(b) y=-2x+6
PEMDAS (order of operations rules) require that we perform operations in a certain order: anything inside parentheses first, followed by any exponentiation, followed by mult. and div., finally followed by addition and subtraction.
Thus, we must evaluate 5+8÷4-1 first, as it's inside parentheses. Focusing on the division first, we get 5+8÷4-1 = 5 + 2 - 1, or 6.
Then we have 5 - [6], which comes out to -1.
Answer:
Draw the triangle.
Draw the perpendicular bisector to each side of the triangle. Draw the lines long enough so that you see a point of intersection of all three lines.
Draw the circle with radius at the intersection point of the bisectors that passes through one of the vertices
Answer:
I think ti is 15 i fi am wrong I am sorry