Answer: 4.8 ft

To answer this question you need to know how much shadow: actual height ratio. Flagpole is having 8 ft shadow with 20 ft actual height. The ratio should be= 8 ft: 20 ft= 0.4

Then multiply the ratio with the oak tree shadow. The equation would be:
0.4 x 12 ft= 4.8 ft

7 0

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:

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