The mean absolute deviation for the number of hours students practiced the violin is 6.4.
<h3>What is the mean absolute deviation?</h3>
The average absolute deviation of the collected data set is the average of absolute deviations from a center point of the data set.
Given
Students reported practicing violin during the last semester for 45, 38, 52, 58, and 42 hours.
The given data set is;
45, 38, 52, 58, 42
Mean Deviation = Σ|x − μ|/N.
μ = mean, and N = total number of values
|x − μ| = |45 − 47| = 2
|38− 47| = 9
|52− 47| = 5
|58− 47| = 11
|42− 47| = 5
The mean absolute deviation for the number of hours students practiced the violin is;
Hence, the mean absolute deviation for the number of hours students practiced the violin is 6.4.
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Answer:
ƒ(x) = (x - 2)^2 + 1
Step-by-step explanation:
To make f(x) be a translation of the graph of g(x) by (h, k), write it as ...
f(x) = g(x -h) +k
You want to translate g(x) = x^2 by (2, 1), 2 units right and 1 unit up, so the function f(x) is ...
f(x) = g(x -2) +1
The rate at which Sean drank the slushy is 5 milliliters per second.
The time it would take to drink all the slushy is 55 seconds.
<h3>How fast did Sean drink?</h3>
In order to determine the speed at which the slushy was drank, divide the slushy drank in 13 seconds by 13 seconds.
Speed = slushy drank in 13 seconds / time
(275 - 210) / 13
65 / 13 = 5 milliliters per second
Time it would take to drink all the slushy =total milliliters of the slushy / speed
275 / 5 = 55 seconds.
To learn more about average speed, please check: brainly.com/question/21734785
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Answer:
Null hypothesis is: U1 - U2 ≤ 0
Alternative hypothesis is U1 - U2 > 0
Step-by-step explanation:
The question involves a comparison of the two types of training given to the salespeople. The requirement is to set up the hypothesis that type A training leads to higher mean weakly sales compared to type B training.
Let U1 = mean sales by type A trainees
Let U2 = mean sales by type B trainees
Therefore, the null hypothesis (H0) is: U1 - U2 ≤ 0
This implies that type A training does not result in higher mean weekly sales than type B training.
The alternative hypothesis (H1) is: U1 - U2 > 0
This implies that type A training indeed results in higher mean weekly sales than type B training.
