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Fudgin [204]
3 years ago
6

Simplify x^2+6x+8/x+2

Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer:

(x + 4)

Step-by-step explanation:

\frac{ {x}^{2}  + 6x + 8}{x + 2}  \\  \frac{ {x}^{2}  + 4x + 2x + 8}{x + 2}  \\  \frac{ x(x + 4) + 2(x + 4) }{(x + 2)}  \\  \frac{(x + 2)(x + 4)}{(x + 2)}  \\  = ( x+ 4)

Hope this helps you.

Let me know if you have any other questions :-)

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Records indicate that for parts coming out a hydraulic repair shop at an airplane rework facility, 30\% will have a shaft defect
mrs_skeptik [129]

Answer:

P(A|B) = P(A∩B)/P(B) = 100%

Which means that there is 100% probability that the item has at least one type of defect given that the item has only a shaft defect.

Step-by-step explanation:

Conditional probability P(A|B) can be expressed as;

P(A|B) = P(A∩B)/P(B) .....1

Given;

30% will have a shaft defect,

15% will have a bushing defect,

and 65% will be defect-free

Total probability = 100% = P(shaft or/and bushing defect) + P(defect free)

P(shaft or/and bushing defect) = 100% - P(defect free)

= 100% - 65% = 35%

And

P(shaft or/and bushing defect) = P(shaft def only) + P(bushing def only) + P(shaft and bushing defect)

P(shaft or/and bushing defect) = P(shaft defect) + P(bushing defect) - P(shaft and bushing defect)

Substituting the values we have;

35% = 30% + 15% - P(shaft and bushing defect)

P(shaft and bushing defect) = 45% - 35% = 10%

Let A="The item has at least one type of defect"; and B="The item has only a shaft defect".

P(A) = P(shaft or/and bushing defect) = 35%

P(B) = P(shaft only defect) = 30% - 10% = 20%

P(A∩B) = 20%

Substituting into equation 1

P(A|B) = P(A∩B)/P(B) = 20%/20%

P(A|B) = 1/1 = 100%

Which means that there is 100% probability that the item has at least one type of defect given that the item has only a shaft defect.

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