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Tcecarenko [31]
3 years ago
6

PLS ANS THIS QUESTION PLS (WITH STEPS) IT WILL BE A GREAT HELP

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

6x² – 10y² + 2xy + 10

Step-by-step explanation:

We'll begin calculating the sum of

x² + 3y² – 6xy and 2x² – y² + 8xy + 8

This can be obtained as follow:

... x² + 3y² – 6xy

+ 2x² – y² + 8xy + 8

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

3x² – 2y² + 2xy + 8

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Next, we shall determine the sum of:

–3x² + 4y² + 3 and 4y² – 5

This can be obtained as follow:

–3x² + 4y² + 3

+ 4y² – 5

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

–3x² + 8y² – 2

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Finally, we shall subtract the sum of:

–3x² + 4y² + 3 and 4y² – 5

from the sum of:

x² + 3y² – 6xy and 2x² – y² + 8xy + 8

This can be obtained as follow:

.. 3x² – 2y² + 2xy + 8

– (–3x² + 8y² – 2)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

6x² – 10y² + 2xy + 10

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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3 years ago
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.

5 0
3 years ago
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