In tossing four fair dice, what is the probability of tossing, at most, one 3

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

                   ( ∵ xxxx )

                   ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

(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