What is the slope of the line defined by the parametric equations x=-2+4t and y=1+4t ?
1 answer:
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Answer:
a) Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.
b) Rachel, because her time had the lowest z-score.
Step-by-step explanation:
Z-score:
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question:
The lower the time, the better the runner is. So whoever's time has a lower z-score is a better runner. So initially, we find the z-score of each student's time.
An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes.
This means that for Rachel's time, we have that . So
A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes.
This means that for Kenji's time, we have that . So
A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.
This means that for Nedda's time, we have that . So
(a) Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?
Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.
(b) Who is the fastest runner with respect to his or her class? Explain why
Rachel, because her time had the lowest z-score.
The distance between two points can be calculated using the following formula:
distance = sqrt [ (x2-x1)^2 + (y2-y1)^2]
Now, we are given the two points (5,4) and (1,-2)
Substitute with the given points in the above equation to get the distance as follows:
distance = sqrt [(1-5)^2 + (-2-4)^2]
distance = sqrt [16+36] = sqrt[52]
distance = 2√13 = 7.2111 units
distance = sqrt [ (x2-x1)^2 + (y2-y1)^2]
Now, we are given the two points (5,4) and (1,-2)
Substitute with the given points in the above equation to get the distance as follows:
distance = sqrt [(1-5)^2 + (-2-4)^2]
distance = sqrt [16+36] = sqrt[52]
distance = 2√13 = 7.2111 units