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balu736 [363]
3 years ago
11

How do you solve for (w-2)²-63=0 where w is a real number

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0
(w-2)^2 = 63

FOIL left hand side:
(w-2)^2 = w^2 - 4w + 4

So,
w^2 - 4w + 4 = 63
w^2 - 4w - 59 = 0

Then use quadratic equation to get values of w = 2 + 3*root(7) and w = 2 - 3*root(7).
±
3
√
7
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Answer:

The slope is

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Step-by-step explanation:

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A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

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M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

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Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

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                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

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Answer:

Totally Rational

Step-by-step explanation:

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<em><u>I hope this helps, dont hesitate to ask for any question.Mark me as brainliest is appreciated.Tq!!</u></em>

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Answer:

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Step-by-step explanation:

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