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:
60 tiles are used
Step-by-step explanation:
1m equals 100 cm.
5m equals 500
3m equals 300
the whole thing is 500 by 300 which when you multiply gives you
150,000. When you multiple 50 by 50 it gives you 2,500. So when you divide 150,000 by 2,500 it gives you a total of 60 tiles. HOPE THIS HELPS
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▹ Answer
<em>2c - 2b - a</em>
▹ Step-by-Step Explanation
(c + b + a) + (-2a - 3b + c)
<u>Remove the parentheses</u>
c + b + a + (-2a - 3b + c)
<u>Collect like terms</u>
2c + b + a - 2a - 3b
<u>Final Answer</u>
2c - 2b - a
Hope this helps!
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Brainliest is greatly appreciated!
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Answer:
4
Step-by-step explanation:
Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:
CW = UCL - LCL
Where,
CW= Class width
UCL= Upper class limit
LCL= Lower class limit
From the table above:
For class 1, CW = 64 - 60 = 4
For class 2, CW = 69 - 65 = 4
For class 3, CW = 74 - 70 = 4
For class 4, CW = 79 - 75 = 4
For class 5, CW = 84 - 80 = 4
Therefore, the class width of the grouped data = 4
