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:
If there are 1000 customers in the store one week, how many will purchase exactly one of these items
1000 CUSTOMERS*28%=280
Step-by-step explanation:
A The event that a persons buys a suit
B The event that a person buys a shirt
C The event that a person buys a tie
P(A)= 22%
P(B)= 30%
P(C)= 28%
P(AB)=
11%
P(AC)=
14%
P(BC)=
10%
P(ABC)=
6%
A u B u C Is the event that any item is bougth
AC u AC u BC Is the event that any two events occured
So the wanted probability is
P[(A u B u C )(AB u AC u BC)^c
P[(A u B u C )=P(AB)+ P(BC)+P(BC)
P[(A u B u C ) =0.22+0.30+0.28-0.11-0.14.-0.10+0.06
=0,51
0,51=+0,23+P[(A u B u C )(AB u AC u BC)^c
=0,28
1000 CUSTOMERS*28%=280
Answer:
k = 575
Step-by-step explanation:
let d be distance and h time.
Given d varies directly as h then the equation relating them is
d = kh ← k is the constant of variation
To find k use the condition d = 2875, h = 5, then
2875 = 5k ( divide both sides by 5 )
k = 575