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.
To know more about mean value click the link given below.
brainly.com/question/5003198
Answer:
Length of truck = 160 +7% of 160 = 160 +11.2 = 171.2inches
Answer:
B. r=7.5
yan po ang correct answer
Check the picture below.
now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.
now, those are the values when the rocket is 30 feet up above.
