Answer:
a) P(X∩Y) = 0.2
b) 
= 0.16
c) P = 0.47
Step-by-step explanation:
Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.
So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67
Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:
P(X∩Y) = P(X) + P(Y) - P(X∪Y)
P(X∩Y) = 0.36 + 0.51 - 0.67
P(X∩Y) = 0.2
On the other hand, the probability that he must stop at the first signal but not at the second one can be calculated as:
= P(X) - P(X∩Y)
= 0.36 - 0.2 = 0.16
At the same way, the probability 
that he must stop at the second signal but not at the first one can be calculated as:
= P(Y) - P(X∩Y)
= 0.51 - 0.2 = 0.31
So, the probability that he must stop at exactly one signal is:
Answer:
The greater than symbols looks like this > , and the less than symbol looks like? <
Answer: <em>m = 7</em>
Step-by-step explanation: In this equation, since a -5 is being multiplied by <em>m</em>, in order to get <em>m</em> by itself, we must divide both sides of the equation by -5.
On the left side, our -5's cancel out and we are left with <em>m</em>. On the right side, -35 ÷ -5 gives us 7. So m = 7.
To check our answer, we plug 7 back in for <em>m</em> in the original equation and we get -5 (7) = -35 which is a true statement so we know our answer is correct.
Don't just do this problem in your head. It's extremely important to develop the habit of putting all your steps down on paper or digitally. It will really pay off for you down the line.
1) in first place you need to Know what kind of figure is it, in order to find solve the problem, as you can see it is a parallelogram, and the condition for this geometric figure the opposites are equal ,
2) if you have a non -rectangular parallelogram and draw a diagonal from A to C, you obtain two equal triangles, for this reason the line AC IS CALLED DIAGONAL, and divide the parallelogram
3) in first place to determine that the angles are congruents for two triangles the Diagonal is comun for both, and the vertex ∡CAB IS THE SAME ∡ACD
4) If the diagonal AB AND CD are equal , AD AND BC are equal too, then the triangles are congruent
5) segment AB Parallel to CD THEY HAVE THE SAME SLOPE AND MAGNITUDE
