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Georgia [21]
3 years ago
7

I have a 78.5% in my history class i have an exam of 52 questions how many questions do i need to get correct to keep a C averag

e or higher?
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0
I think you'd have to get 41 questions right . But it also depends on the grading scale how many points do you get a question?

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Write a trinomial that has a factor of X +3 and a GCF of -5x
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Answer:

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Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the nu
olga_2 [115]

Answer:

a

The probability that the selected joint was judged to be defective by neither of the two inspectors is   P(A' n B' ) = 0.8855

b

The probability that the selected joint was judged to be defective by inspector B but not by inspector A  is  P(A' n B) =0.0403

Step-by-step explanation:

From the question we are told that

   The sample size is n_s =  10000

    The number of outcome for inspector A is  n__{A}} = 742

    The number of outcome for inspector B is  n__{B}} = 745

     The number of joints judged defective by both inspector is n(A u B) =  1145

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

      P(A' n B' ) =  1 - P(A u B)

Now

       P(A\ u \ B) = \frac{n(Au B)}{n_s }

substituting values

        P(A\ u \ B) = \frac{1145}{ 10 000 }

So  

      P(A' n B' ) =  1 - \frac{1145}{10 000}

     P(A' n B' ) = 0.8855

the probability that the selected joint was judged to be defective by inspector B but not by inspector A  is mathematically represented as

     P(A' n B) =  P(A \ u \ B) -P(A)

Now

        P(A) =  \frac{n__{A}}{n_s}

substituting values

       P(A) =  \frac{742}{10 000}

So

     P(A' n B) =   \frac{1145}{10 000}  - \frac{742}{10 000}

    P(A' n B) =0.0403

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Combine the second

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Then you should come up with the sloution 2x - 2y

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