Let
x = the number of a-cables
y = the number of b-cables
z = the number of c-cables
Cable a uses 3 black, 3 white and 2 red wires.
Cable b uses 1 black, 2 white and 1 red wire.
Cable c uses 2 black, 1 white and 2 red wires.
Because 120 black, 110 white and 100 red wires were used, therefore the system of equations is
3x + y + 2z = 120 (1)
3x + 2y + z = 110 (2)
2x + y + 2z = 100 (3)
Answer:
3x + y + 2z = 120
3x + 2y + z = 110
2x + y + 2z = 100
where
x,y,z = the number of a, b and c-cables respectively.
Answer: P(Upper E' prime) = 0.06
Step-by-step explanation:
Not happen:P(Upper E prime) = 0.94
Happen: P(Upper E' prime) = ?
P(Upper E' prime) + P(Upper E' prime) = 1
Therefore
P(Upper E' prime) + 0.94= 1
P(Upper E' prime) =1- 0.94
P(Upper E' prime) = 0.06
Answer:
y=-4 x=1
Step-by-step explanation:
y = 1/2x + 3
x - y = 5
x-1/2x + 3=5
-3 -3
1/2x=2
1/2 1/2
x=1
1-y=5
-1 -1
-y=4
/-1 /-1
y=-4
Step-by-step explanation:
2-5/4x=13
- (8x-5)/4x=13
- 8x-5=52x
- -5=52x-8x
- -5=44x
- x=-5/44
9514 1404 393
Answer:
multiply by 2/√3
Step-by-step explanation:
The ratio of the hypotenuse to the longer leg is 2/√3. The hypotenuse can be found by multiplying the length of the longer leg by 2/√3.