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scoray [572]
2 years ago
7

X-9/3=56/x+4 solve for x

Mathematics
1 answer:
Flauer [41]2 years ago
7 0

Answer:

x = sqrt(273)/2 - 1/2 or x = -1/2 - sqrt(273)/2

Step-by-step explanation:

Solve for x:

x - 3 = 56/(x + 4)

Cross multiply:

(x - 3) (x + 4) = 56

Expand out terms of the left hand side:

x^2 + x - 12 = 56

Add 12 to both sides:

x^2 + x = 68

Add 1/4 to both sides:

x^2 + x + 1/4 = 273/4

Write the left hand side as a square:

(x + 1/2)^2 = 273/4

Take the square root of both sides:

x + 1/2 = sqrt(273)/2 or x + 1/2 = -sqrt(273)/2

Subtract 1/2 from both sides:

x = sqrt(273)/2 - 1/2 or x + 1/2 = -sqrt(273)/2

Subtract 1/2 from both sides:

Answer:  x = sqrt(273)/2 - 1/2 or x = -1/2 - sqrt(273)/2

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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
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Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

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We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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