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Aliun [14]
2 years ago
11

Runner A crosses the starting line of a marathon and runs at an average pace of 5.6 miles per hour. Half an hour later, Runner B

crosses the starting line and runs at an average rate of 6.4 miles per hour. If the length of the marathon is 26.2 miles, which runner will finish ahead of the other? Explain.
A.
Runner B; Runner B will pass Runner A and finish more than half an hour ahead of Runner A.


B.
Runner B; Runner B will catch up to Runner A 4 hours after Runner A crosses the starting line.


C.
Runner A; Runner B will not be able to catch Runner A in the time it takes Runner A to complete the 26.2 mile course.


D.
Runner B; Runner B will catch up to Runner A 3.5 hours after Runner A crosses the starting line.

It's on Gradpoint
Mathematics
1 answer:
makvit [3.9K]2 years ago
3 0

Answer:

D. Runner B; Runner B will catch up to Runner A 3.5 hours after Runner A crosses the starting line.

Step-by-step explanation:

Runner A:

Time it takes for him to finish:

t = 26.2/5.6 = 4.679 hours (to complete race)

.

Runner B:

Time it takes for him to finish:

t = 26.2/6.4 = 4.094 hours

BUT wait, we have to account for the extra .5 hours -- the time it took for him to cross the start line.

t = 4.094 + .5 = 4.594 hours

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Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

Hence,

The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

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3 years ago
Mari spends $6 on cheese for a party she also plans to buy b boxes of crackers that cost $3 per box
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Answer:

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Researchers studied symptom distress and palliative care designation among a sample of 710 hospitalized patients. Controlling fo
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A. Men had less distress from nausea on average than women but we can not determine if this is a significant difference.

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Working based on the information given, the mean values of each group with with men having an average score of 1.02 and women have an average of 1.79 this reveals that distressing nausea on average is higher in women than in men . However, to test if there is a significant difference would be challenging as the information given isn't enough to make proceed with the test as the standard deviations of the two groups aren't given and no accompanying sample data is given.

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Answer:

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3 years ago
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