Six people, including a, b, and c, form a queue in a random order (all 6! orderings are equiprobable). consider the event "b is between a and c in the queue". what is its probability? (the order of a and c can be arbitrary, but b should be between them)

There is a 1.39% probability that "B is between A and C in the queue".

Step-by-step explanation:

The first step to solve this problem is find the total number of possible orderings:

The first person of the queue can be any of the six. The second, can be any but the first, so five.

So

There are 6*5*4*3*2*1 = 720 total queue orderings.

Now we find the number of queues that B is between A and C. So:

We have:

B as the second, A as the first and C as the third

B as the second, C as the first and A as the third

B as the third, A as the second and C as the fourth

B as the third, C as the second and A as the fourth

B as the fourth, A as the third and C as the fifth.

B as the fourth, C as the third and A as the fifth.

B as the fifth, A as the fourth and C as the sixth.

B as the fifth, C as the fourth and A as the sixth.

So 8 total outcomes in which B is between A and C.

What is its probability?

There is a 1.11% probability that "B is between A and C in the queue".

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116