Probability of a standard six side dice landing on a four in the first three rolls?

91/216

Step-by-step explanation:

The probability of getting a 4 in the first three rolls is 1 minus the probability of not getting a 4 on any of the rolls.

P(at least one 4) = 1 − P(no 4s)

P(at least one 4) = 1 − (5/6)³

P(at least one 4) = 91/216

Alternatively, you can calculate it this way.

The probability of getting a 4 on the first roll is 1/6.

The probability of getting a 4 on the second roll is (5/6) (1/6) = 5/36.

The probability of getting a 4 on the third roll is (5/6) (5/6) (1/6) = 25/216.

The probability of any of the three events is 1/6 + 5/36 + 25/216 = 91/216.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3