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vovangra [49]
2 years ago
7

Presenting someone else's words or ideas as your own is:

Mathematics
2 answers:
mote1985 [20]2 years ago
7 0

Answer:

<em>B. plagiarizing</em>

Step-by-step explanation:

...........

Mumz [18]2 years ago
6 0
B plagiarizing because it means to copy someone.
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charle [14.2K]
The answer is 16 because (-2*-2*-2*-2= 16)
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What is the unit rate for the proportional relationship represented by the equation y=4/5x
slava [35]

Answer:

<em>the unit rate is 0.8x per unit y</em>

<em></em>

Step-by-step explanation:

the proportional relationship is y = \frac{4x}{5}

this means that the constant of proportionality k = \frac{4}{5} = 0.8

re-writing, we have

<em>y = 0.8x</em>

<em>therefore, the unit rate is 0.8x per unit y</em>

5 0
3 years ago
(4x+3)(3x)-25 graph
maw [93]

Answer:

12x^2+9x^2-25    (quadratic equation)

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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zavuch27 [327]

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6 0
3 years ago
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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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