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4 years ago

Completely factor 12x² - 7x - 12.

1 answer:

4 0

You can solve this a few ways. You could use the quadratic formula but I am going to use the grouping method.

$12x^2 - 7x - 12$
First find two numbers that equal -12 when you multiply them but also equals -7 when you add them. Two numbers that do just that are -16 and 9

Now use the two numbers and put it in group form
$12x^2 - 7x - 12$
$12x^2 - 16x + 9x - 12$
$(12x^2 - 16x) + (9x - 12)$

Now factor out the groups
$(12x^2 - 16x) + (9x - 12)$
$4x(3x - 4) + 3(3x - 4)$
$(4x + 3)(3x - 4)$
The answer is A

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hammer [34]

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:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago