1. solve, using the substitution method.4x + 2y = 11x - 2 = - 2ya. ( 3 , - 1/2 )b. ( 0, -2 )c. ( 0, - 11/2 )d. ( 7, - 5/2 )2. solve, using the substitution method. y= 3x + 54x - y = 5a. ( 10, 35 )b. ( 15, 10 )c. there are an infinite number of solutions. d. there is no solution.

1. 4x + 2y = 11

x - 2 = -2y

First I would isolate one of the variables (x or y) of one of the equations, and then substitute it into the other equation.

The easiest to isolate is the "x" in the second equation

x - 2 = -2y  Add 2 on both sides

x = -2y + 2

Substitute this into the first equation

4x + 2y = 11

4(-2y + 2) + 2y = 11   Multiply 4 into (-2y + 2)

-8y + 8 + 2y = 11  Combine like terms

-6y + 8 = 11  Subtract 8 on both sides

-6y = 3   Divide -6 on both sides

y = -3/6  Simplify

y = -1/2

Now that you know "y", you can plug it into either of the original equations to find "x"

x - 2 = -2y

x - 2 = -2(-1/2)

x - 2 = 1    Add 2 on both sides

x = 3

Answer is A

2. y = 3x + 5

4x - y = 5

Substitute the first equation into the second equation

4x - y = 5

4x - (3x + 5) = 5   Multiply/distribute the - into (3x + 5)

4x - 3x - 5 = 5  Combine like terms

x - 5 = 5  Add 5 on both sides

x = 10

Plug in "x" into either of the original equations to find "y"

y = 3x + 5

y = 3(10) + 5

y = 30 + 5

y = 35

Answer is A

differnt varibles efect the range

step-by-step explanation:

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0