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fredd [130]
3 years ago
15

Please help with the question in the picture!!!

Mathematics
1 answer:
Sindrei [870]3 years ago
6 0

Answer:

We have the measures:

1023 cm

2.3 m

8.72m

6430 mm

1200 mm

6.4 cm

2.5m

0.06km

Now let's write all those measures in the same unit, let's use meters.

We know that:

1cm = 0.01m

1mm = 0.001m

1 km = 1000m

Then, we can rewrie the measures as:

1023 cm = 1023*0.01 m = 10.23 m

2.3 m

8.72m

6430 mm = 6430*0.001 m = 6.430 m

1200mm = 1200*0.001 m = 1.2 m

6.4 cm = 6.4*0.01m =  0.064 m

2.5m

0.06km = 0.06*1000km = 60m

Then the order, from smallest to largest is:

6.4 cm

1200mm

2.3 m

2.5m

6430 mm

8.72m

1023 cm

0.06km

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Answer:

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Both f(x) and h(x) have the same (greatest) maximum value.

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3 years ago
2x<15 solve for x
Nadusha1986 [10]

Answer:

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3 years ago
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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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.

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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)

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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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