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Dmitriy789 [7]
3 years ago
13

Samara has an assignment to conduct a survey. She wants to find out what students like to do after school. She asks the first 10

people she sees in her Book Club what they like to do after school. Which statement correctly explains whether Samara's sample is a random sample?
A.Samara’s sample is not random because she only asks 10 people.
B.Samara's sample is random because she asks each person the same question.
C.Samara’s sample is random because she asks the first 10 people she sees.
D.Samara’s sample is not random because she only asks people in a book club
Mathematics
2 answers:
Sliva [168]3 years ago
7 0
C: Because she doesn't choose certain people
Romashka-Z-Leto [24]3 years ago
6 0
Answer D because if she wanted real results, then she would ask 10 people she saw in the whole school instead of just the book club
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Answer:

355,459.33

Step-by-step explanation:

42,655.12/12 = 3554.59

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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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Answer:

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Answer:

Step-by-step explanation:

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