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Andrej [43]
3 years ago
12

Herman and his dad are leaving early in the morning for his lacrosse tournament. Their house is 210 miles from the tournament. T

hey plan to stop and eat after 2 hours of driving, then compete the rest of the trip. Herman's dad plans to drive at an average speed of 60 miles per hour. How long will the second part of the trip take
Mathematics
1 answer:
Licemer1 [7]3 years ago
6 0

The second part of the trip took 1.5 hours.

Average Speed is the ratio of the total distance travelled to the total time taken to cover that distance. It is given by:

Average Speed = total distance ÷ total time

Given that the distance from the house to the tournament is 210 miles, hence:

Total distance = 210 miles

Also, The journey was in two parts, the first part took 2 hours. Let the second part take t hours. Therefore:

Total time = t + 2

The average speed of 60 miles per hour, hence:

60 = 210 ÷ (t + 2)

60(t + 2) = 210

60t + 120 = 210

60t = 90

t = 1.5 hours

Therefore the second part of the trip took 1.5 hours.

Find out more at: brainly.com/question/9834403

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a. 0.9931

b. 0.3423

c. 0.3907

d. 0.2670

e. 3.15

f. 1.0796

Step-by-step explanation:

The probability of the variable that said the number of left-handed people follow a Binomial distribution, so the probability is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

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nCx=\frac{n!}{x!(n-x)!}

Where x is the number of left-handed people, n is the number of people selected at random and p is the probability that a person is left--handed. So P(x) is:

P(x)=5Cx*0.63^{x}*(1-0.63)^{5-x}

Then the probabilities P(0), P(1), P(2), P(3), P(4) and P(5) are:

P(0)=5C0*0.63^{0}*(1-0.63)^{5-0}=0.0069

P(1)=5C1*0.63^{1}*(1-0.63)^{5-1}=0.0590

P(2)=5C2*0.63^{2}*(1-0.63)^{5-2}=0.2011

P(3)=5C3*0.63^{3}*(1-0.63)^{5-3}=0.3423

P(x)=5C4*0.63^{4}*(1-0.63)^{5-4}=0.2914

P(x)=5C5*0.63^{5}*(1-0.63)^{5-5}=0.0993

Then, the probability P(x≥1) that there are some lefties among the 5 people is:

P(x≥1) = P(1) + P(2) + P(3) + P(4) + P(5)

P(x≥1) = 0.0590 + 0.2011 + 0.3423 + 0.2914 + 0.0993 = 0.9931

The probability P(3) that there are exactly 3 lefties in the group is:

P(3) = 0.3423

The probability P(x≥4) that there are at least 4 lefties in the group is:

P(x≥4) = P(4) + P(5) = 0.2914 + 0.0993 = 0.3907

The probability P(x≤2) that there are no more than 2 lefties in the group is:

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E(x)=np=5(0.63)=3.15\\S(x)=\sqrt{np(1-p)}=\sqrt{5(0.63)(1-0.63)}=1.0796

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