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Salsk061 [2.6K]
3 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.
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.
Runner B; Runner B will catch up to runner A 3.5 hours after Runner A crosses the starting line.
Runner B; Runner B will pass Runner A and finish more than half an hour ahead of Runner A.
Runner B; Runner B will catch up to Runner A 4 hours after Runner A crosses the starting line.
Mathematics
1 answer:
marishachu [46]3 years ago
5 0
Time it took runner A to complete the marathon = 26.2 / 5.6 = 4 hrs 41 mins
Time it took runner B to complete the marathon = 26.2 / 6.4 = 4 hrs 6 mins

Time it took runner B to complete the marathon relative to when runner A started = 30 mins + 4 hrs 6 mins = 4 hrs 36 mins

Therefore, runner B will finnish ahead of runner A.

Let x be the time the two runners are at the same point, then
5.6x + 5.6(0.5) = 6.4x
6.4x - 5.6x = 2.8
0.8x = 2.8
x = 2.8/0.8 = 3.5
Therefore, runner B will catch up with runner A 3.5 hours after runner A starts the race.
<span>Runner B; Runner B will catch up to Runner A 3.5 hours after Runner A crosses the starting line.</span>

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a) P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

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Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

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Where \mu=25 and \sigma=5

Part a

For this case we want to find this probability:

P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

And if we use the z score given by:

z =\frac{x-\mu}{\sigma}

We got this:

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

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Part b

For this case we want this probability:

P(X \leq 30)

And if we use the continuity correction we got:

P(X \leq 30)= P(X

And using the z score we got:

P(X

Part c

For this case we want this probability:

P(X

And if we use the continuity correction we got:

P(X

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