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Digiron [165]
1 year ago
15

What is x. Please show *all* the steps.

Mathematics
1 answer:
Zolol [24]1 year ago
8 0

The equation 5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0 is a quadratic equation

The value of x is 8 or 1

<h3>How to determine the value of x?</h3>

The equation is given as:

5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Rewrite as:

-5/x - 2 + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Take the  LCM

[-5(x + 2) + (x -5)(x- 2)]\[x^2 - 4 + [3x + 8]/[x^2 - 4] = 0

Expand

[-5x - 10 + x^2 - 7x + 10]/[x^2 - 4] + [3x + 8]/[x^2 - 4] = 0

Evaluate the like terms

[x^2 - 12x]/[x^2 - 4] + [3x + 8]/[x^2 - 4 = 0

Multiply through by x^2 - 4

x^2 - 12x+ 3x + 8 = 0

Evaluate the like terms

x^2 -9x + 8 = 0

Expand

x^2 -x - 8x + 8 = 0

Factorize

x(x -1) - 8(x - 1) = 0

Factor out x - 1

(x -8)(x - 1) = 0

Solve for x

x = 8 or x = 1

Hence, the value of x is 8 or 1

Read more about equations at:

brainly.com/question/2972832

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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:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

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

                = [0.20]⁴

                = 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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Factor −25x−25

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