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azamat
2 years ago
6

The table below shows some inputs and outputs of the invertible function f ff with domain all real numbers.

Mathematics
1 answer:
Marina CMI [18]2 years ago
6 0

Answer: f^{-1}(1)+f(-14)=20

f^{-1}(-2)=10  

Step-by-step explanation:

The given table :

x: -14,-7,-12,9,10,-2

f(x):11,-12,5,1,-2,13

Since f is invertible ( given) , then f^{-1}(x)  exists.

Now , from table f^{-1}(1)=9  [ x= 9 corresponding to f(x) =1]

f(-14)=11                            [ f(x) = 11 corresponding to x=-14]

then, f^{-1}(1)+f(-14)=9+11=20

So, f^{-1}(1)+f(-14)=20

Also, x= 10 corresponding to f(x) =-2, then

f^{-1}(-2)=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%

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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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