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:
(-6,5)
Step-by-step explanation:
Answer:
18
Step-by-step explanation:
xy + 4 = 40
xy = 40 - 4
xy = 36
What two numbers can b multiplied to give 36? Then add them.
9 + 4 = 13
18 + 2 = 20
6 + 6 = 12
12 + 3 = 15
The odd option is 18
Answer:
Solution : (15, - 11)
Step-by-step explanation:
We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )
Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )
Row Echelon Form :
Step # 1 : Swap the first and second matrix rows,
Step # 2 : Cancel leading coefficient in row 2 through ,
Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.
As you can see our solution is x = 15, y = - 11 or (15, - 11).