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yanalaym [24]
2 years ago
11

What is the solution of radical x+12 equal x

Mathematics
2 answers:
Vlad [161]2 years ago
6 0

Answer:

x = 4 or -3

Step-by-step explanation:

\sqrt{x+12} = x

To get rid of the square root, we need to square both sides.

x + 12 = x²

Get all variables/ integer on to one side.

0 = x² - x - 12

Use any method to solve this quadratic equation. I'm going to use the completing the square method.

12 = x² - x

12 + 0.25 = x² - x + 0.25

\sqrt{12.25} = \sqrt{(x - 0.5)^2}

±3.5 = x - 0.5

x = 0.5 ± 3.5

x = 4 or -3

DIA [1.3K]2 years ago
3 0
Sqrt(x+12)=x
*square root both sides
x+12=x^2
*subtract over x and 12
x^2-x-12=0
*factor using diamond method
(x-4)(x+3)=0
x=4,-3
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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
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Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

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(d) The probability that at most 1 of the next three vehicles passes is 0.216.

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Step-by-step explanation:

Let <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

                                    =(0.70)^{3}\\=0.343

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

                                   =1-0.343\\=0.657

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

                         = P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

                              + P (Only 3rd vehicle passes)

                       =(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

                                                       + P (0 vehicles passes)

                                              =0.189+(0.30\times0.30\times0.30)\\=0.216

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

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