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matrenka [14]
3 years ago
13

What is the Y-intercept in the equation y= 4x -4?

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

The y-intercept is -4

Step-by-step explanation:

You have your x intercept, which is 4 and your y is -4. Hope this helps :)

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At Emily’s closet 1/3 of the cloths are white and 1/5 are black what. What fraction of Emilly’s clothes are black and white ?
vfiekz [6]

8/15

  1. Since 1/5 and 1/3 have different denominators, they cannot be added directly.
  2. Multiply the denominators (5×3=15)
  3. Multiply each with the other denominator 3×(1/5)=3/15, 5×(1/3)=5/15
  4. Now, 3/15 +5/15=8/15
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3 years ago
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39 students have a pet because 130 times .30 which represents thirty percent gives you the answer

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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.

6 0
3 years ago
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