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saul85 [17]
3 years ago
15

A fair number cube marked 1, 2, 3, 4, 5, and 6 is rolled. What is the chance that an even number will be rolled?

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
8 0

Answer:

50% chance

Step-by-step explanation:

There are a total of 6 answers, 3 of them are odd and 3 are even. Half of six is  3 so the fraction is 3/6 which can be simplified to 1/2 or .50 or 50%.

olchik [2.2K]3 years ago
6 0

Answer:

50% or 1/2

Step-by-step explanation:

Chance is like probability, so the chance of rolling a even number out of  1, 2, 3, 4, 5, and 6 is half of them, they are 2, 4, and 6. In fraction form it is 1/2, and in percent form it is %0%. Hope this helps.

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Which of the following sequences are not geometric? (check all that apply) a. 2,10,50,250,1250 b. 1,4,9,16,25,36 c. -4,-2,-1,-0.
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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

7 0
3 years ago
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