Answer:
The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
Step-by-step explanation:
The three different assembly lines are: A₁, A₂ and A₃.
Denote <em>R</em> as the event that a component needs rework.
It is given that:
Compute the probability that a randomly selected component needs rework as follows:
Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:
Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.
The answer is:
4x+2y=11} 4x+2y=11} x=2-2y
x-2=-2y } x+2y=2 } x=2-2(-3/6)
x=2-(-6/6)
4(2-2y)+2y=11 x=2-(-1)
8-8y+2y=11 x=3
8-6y=11
-6y=11-8
-6y=3/-6
y=-3/6=-1/2
So for my opinion the solution is A.(3,-1/2)
The diameter is twice as big as the radius
In order to divide polynomials using synthetic division, you must be dividing by a linear expression and the leading coefficient (first number) must be a 1. For example, you can use synthetic division to divide by x + 3 or x – 6, but you cannot use synthetic division to divide by x2 + 2 or 3x2 – x + 7. If the leading coefficient is not a 1, then you must divide by the leading coefficient to turn the leading coefficient into a 1. For example, 3x – 1 would becomex minus 1/3 and 2x + 7 would becomex plus 7/2. If synthetic division will not work, then you must use long division.
Answer:
can yhu see the answer i put?
Step-by-step explanation:
