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Ratling [72]
3 years ago
9

Suppose an experiment has 3 stages: A, B, and C. If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes.

how many outcomes does the entire experiment have?
Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Answer:

72

Step-by-step explanation:

Stage A has 6 outcomes, then B has 4 and C has 3

Take for example only 1 outcome of A, it will have 4 outcomes of B and each outcome of B will have 3 outcomes.

Then a single outcome of A has a total of 12 outcomes (4*3)

Writing it in a way A, B, C:

1,1,1

1,1,2

1,1,3

1,2,1

1,2,2

1,2,3

1,3,1

1,3,2

1,3,3

1,4,1

1,4,2

1,4,3

This is only taking in consideration 1 outcome of A, A has a total of 6 outcomes.

The total of outcomes in the experiment = 6*12=72

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<h3>            x = -9,  y = -13 </h3><h3>    or    x = 13,   y = 9</h3><h3>    or    x = -13,  y = -9</h3><h3>    or     x = 9,   y = 13</h3>

Step-by-step explanation:

x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad  x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}

y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9

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Hope it helps!

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