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Pavel [41]
3 years ago
14

Why is it best to compare medians and interquartile ranges for these​ data, rather than comparing means and standard​ deviations

? A. Some of these data have vastly different​ centers, and the median and interquartile range provide more reliable information in these circumstances. B. These data are fairly symmetric and do not have​ outliers, and the median and interquartile range provide more reliable information in these circumstances. C. Some of these data have outliers​ and/or are​ skewed, and the median and interquartile range are resistant to outliers. D. Some of these data have outliers​ and/or are​ skewed, and the median and interquartile range amplify the effects of outliers and skewness.
Mathematics
1 answer:
Elanso [62]3 years ago
8 0

Answer:Some of these data have outliers​ and/or are​ skewed, and the median and interquartile range are resistant to outliers.

Step-by-step explanation:

Outliers often characterize a lot of data. Outliers are extreme values which have obvious impact on the value of the standard deviation of a given distribution.

However, the median and interquartile range are resistant to outliers hence they can be used even in the presence of outliers in the distribution.

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<em>Areas under the graphs:</em>

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<u />(1)(0.5)+(7-2)(0.2)=1.5\\\\

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As neither of these graphs have an area of 1, neither of them are density curves.

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Fill in the blanks below in order to justify whether or not the mapping shown
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Mrs. Carter baked 100 muffins for a bake sale. The muffins were sold in packages in 2. There were 12 muffins left. Write an equa
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Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia
ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account  the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

              Xa = Xb

     -6 + (3*t) =  7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

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                  t = 13.4/6.8

                  t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0
3 years ago
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