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Montano1993 [528]
1 year ago
12

an urn contains 5 blue balls and 7 orange balls. in how many ways can we select 3 blue balls and 5 orange balls from the urn?

Mathematics
1 answer:
Temka [501]1 year ago
3 0

The no. of ways to select the 3 blue balls and 5 orange balls from the urn is 0.424

No. of Blue balls in the urn = 5

No. of Orange balls in the urn = 7

Total no. of balls in the urn = No. of blue balls + No. of orange balls

                                             = 5 + 7

                                            = 12 balls

No. of blue balls to be selected = 3

No. of orange balls to be selected = 5

No. of balls to be selected = 8

The total. no of ways 8 balls can be selected = 12C₈

The No. of ways 3 blue balls can be selected = 5C₃

The No. of ways 5 orange balls can be selected = 7C₅

Therefore, No. of ways to select 3 blue balls and 5 orange balls from the urn

                       =  5C₃ x  7C₅ / 12C₈

[ Probability = No. of ways to select blue balls x No. of ways to select orange balls / Total no. of ways to select balls]

               

                P = 10 x 21 / 495

                   = 210 / 495

                P = 0.424

Therefore,  the no. of ways to select blue and orange balls is 0.424

Learn more about the probability in

brainly.com/question/11234923

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Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

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By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

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Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

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Therefore, the solution of the given differential equation is

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