The correct answer is 4> -3
I think your answer is gonna be 20
I believe the answer is True
Answer:
1. D. 20, 30, and 50
2. A. 86
3. B. 94
Step-by-step explanation:
1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.
The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.
Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.
Thus:
Q1 = (85 + 87)/2 = 86
Q3 = (93 + 95)/2 = 94
IQR = Q3 - Q1 = 94 - 86
IQR = 8
Outliers in the data set are data values below the lower limit or above the upper limit.
Let's find the lower and upper limit.
Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74
The data values below the lower limit (74) are 20, 30, and 50
Let's see if we have any data value above the upper limit.
Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106
No data value is above 106.
Therefore, the only outliers of the data set are:
D. 20, 30, and 50
2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.
Thus:
Q1 = (85 + 87)/2 = 86
3. Q3 = (93 + 95)/2 = 94
m x H = ![\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2637.5%26-12.5%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Step 1; Multiply 5 with this matrix
and we get a matrix ![\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%2610%5C%5C20%2640%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiply the fraction
with the matrix
and we get ![\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2m%7D%7B5%7D%20%26%5Cfrac%7B4m%7D%7B5%7D%20%5C%5C%5Cfrac%7B8m%7D%7B5%7D%20%26%5Cfrac%7B16m%7D%7B5%7D%20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step2; Now equate corresponding values of the matrices with each other.
-5 =
and so on. By equating we get the value of m as 
Step 3; Add the matrices to get the value of matrix m.
Adding the three matrices on the RHS we get
.
Step 4; Adding the matrices on the LHS we get the resulting matrix as H +
. Equating the matrices from step 3 and 4 we get the value of H as ![\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%26-1%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step 5; Now to find the value of m x H we need to multiply the value of
with the matrix
Step 6; Multiplying we get the matrix m x H = [ -25
]