Answer :
Let's break down the problem into smaller parts and solve it step-by-step.
### Part 1: Compute the Mean and Range
#### Mean
Given the data: [tex]\(14, 22, 20, 7, 13, 14, 8\)[/tex], we calculate the mean (average) by summing the values and dividing by the number of values.
[tex]\[ \text{Mean} = \frac{14 + 22 + 20 + 7 + 13 + 14 + 8}{7} \][/tex]
[tex]\[ \text{Mean} = \frac{98}{7} = 14 \][/tex]
#### Range
Given the data: [tex]\(4, 11, 12, 8, 9, 10, 15, 19, 25\)[/tex], the range is the difference between the maximum and minimum values.
[tex]\[ \text{Range} = \max(4, 11, 12, 8, 9, 10, 15, 19, 25) - \min(4, 11, 12, 8, 9, 10, 15, 19, 25) \][/tex]
[tex]\[ \text{Range} = 25 - 4 = 21 \][/tex]
### Part 2: Solving the System of Matrices
Given the equation:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} : \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
Here, it appears that ':' might imply matrix multiplication. Let's treat this as finding values [tex]\(x\)[/tex] and [tex]\(y\)[/tex] such that:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
Let's perform the multiplication:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 2x \\ 6 & 8 \end{bmatrix} \][/tex]
Multiplying with the second matrix:
[tex]\[ \begin{bmatrix} 2 & 2x \\ 6 & 8 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 2 \cdot 2 + 2x \cdot y & 2 \cdot 1 + 2x \cdot 3 \\ 6 \cdot 2 + 8 \cdot y & 6 \cdot 1 + 8 \cdot 3 \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 4 + 2xy & 2 + 6x \\ 12 + 8y & 6 + 24 \end{bmatrix} \][/tex]
Equating this to the given matrix:
[tex]\[ \begin{bmatrix} 4 + 2xy & 2 + 6x \\ 12 + 8y & 30 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
We solve these equations:
1. [tex]\( 4 + 2xy = 0 \Rightarrow 2xy = -4 \Rightarrow xy = -2 \)[/tex]
2. [tex]\( 2 + 6x = 7 \Rightarrow 6x = 5 \Rightarrow x = \frac{5}{6} \)[/tex]
3. [tex]\( 12 + 8y = 6 \Rightarrow 8y = -6 \Rightarrow y = -\frac{3}{4} \)[/tex]
4. [tex]\( 30 = 30 \)[/tex] is trivially true.
So, the values are [tex]\( x = \frac{5}{6} \)[/tex] and [tex]\( y = -\frac{3}{4} \)[/tex].
### Part 3: Finding the Domain of the Function
Given the function [tex]\( f(x) = 4x + 7 \)[/tex] and the range [tex]\(\{311, 15, 19\}\)[/tex], we find the domain by solving each value in the range for [tex]\(x\)[/tex]:
1. For [tex]\(311\)[/tex]:
[tex]\[ 311 = 4x + 7 \Rightarrow 4x = 304 \Rightarrow x = 76 \][/tex]
2. For [tex]\(15\)[/tex]:
[tex]\[ 15 = 4x + 7 \Rightarrow 4x = 8 \Rightarrow x = 2 \][/tex]
3. For [tex]\(19\)[/tex]:
[tex]\[ 19 = 4x + 7 \Rightarrow 4x = 12 \Rightarrow x = 3 \][/tex]
So the domain is [tex]\(\{76, 2, 3\}\)[/tex].
### Part 4: Creating the Arrow Diagram
Given:
[tex]\[ M = \{e^8, 2\}, \quad N = \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
To find [tex]\( MM \times N \)[/tex]:
[tex]\[ M = \begin{pmatrix} e^8 \\ 2 \end{pmatrix} \][/tex]
[tex]\[ N = \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
We need to form the outer product:
[tex]\[ MM \times N = \begin{pmatrix} e^8 \\ 2 \end{pmatrix} \times \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
The resulting matrix is:
[tex]\[ \left[ \begin{array}{cc} e^8 \cdot a & e^8 \cdot b \\ 2 \cdot y & 2 \cdot 2 \end{array} \right] \][/tex]
[tex]\[ = \begin{pmatrix} e^8 a & e^8 b \\ 2y & 4 \end{pmatrix} \][/tex]
This is the outer product matrix.
In summary:
- Mean = 14
- Range = 21
- Values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]: [tex]\( x = \frac{5}{6}, y = -\frac{3}{4} \)[/tex]
- Domain = [tex]\(\{76, 2, 3\}\)[/tex]
- Arrow Diagram Result: [tex]\(\begin{pmatrix} e^8 a & e^8 b \\ 2y & 4 \end{pmatrix}\)[/tex]
### Part 1: Compute the Mean and Range
#### Mean
Given the data: [tex]\(14, 22, 20, 7, 13, 14, 8\)[/tex], we calculate the mean (average) by summing the values and dividing by the number of values.
[tex]\[ \text{Mean} = \frac{14 + 22 + 20 + 7 + 13 + 14 + 8}{7} \][/tex]
[tex]\[ \text{Mean} = \frac{98}{7} = 14 \][/tex]
#### Range
Given the data: [tex]\(4, 11, 12, 8, 9, 10, 15, 19, 25\)[/tex], the range is the difference between the maximum and minimum values.
[tex]\[ \text{Range} = \max(4, 11, 12, 8, 9, 10, 15, 19, 25) - \min(4, 11, 12, 8, 9, 10, 15, 19, 25) \][/tex]
[tex]\[ \text{Range} = 25 - 4 = 21 \][/tex]
### Part 2: Solving the System of Matrices
Given the equation:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} : \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
Here, it appears that ':' might imply matrix multiplication. Let's treat this as finding values [tex]\(x\)[/tex] and [tex]\(y\)[/tex] such that:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
Let's perform the multiplication:
[tex]\[ 2 \begin{bmatrix} 1 & x \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 2x \\ 6 & 8 \end{bmatrix} \][/tex]
Multiplying with the second matrix:
[tex]\[ \begin{bmatrix} 2 & 2x \\ 6 & 8 \end{bmatrix} \cdot \begin{bmatrix} 2 & 1 \\ y & 3 \end{bmatrix} = \begin{bmatrix} 2 \cdot 2 + 2x \cdot y & 2 \cdot 1 + 2x \cdot 3 \\ 6 \cdot 2 + 8 \cdot y & 6 \cdot 1 + 8 \cdot 3 \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 4 + 2xy & 2 + 6x \\ 12 + 8y & 6 + 24 \end{bmatrix} \][/tex]
Equating this to the given matrix:
[tex]\[ \begin{bmatrix} 4 + 2xy & 2 + 6x \\ 12 + 8y & 30 \end{bmatrix} = \begin{bmatrix} 0 & 7 \\ 6 & 5 \end{bmatrix} \][/tex]
We solve these equations:
1. [tex]\( 4 + 2xy = 0 \Rightarrow 2xy = -4 \Rightarrow xy = -2 \)[/tex]
2. [tex]\( 2 + 6x = 7 \Rightarrow 6x = 5 \Rightarrow x = \frac{5}{6} \)[/tex]
3. [tex]\( 12 + 8y = 6 \Rightarrow 8y = -6 \Rightarrow y = -\frac{3}{4} \)[/tex]
4. [tex]\( 30 = 30 \)[/tex] is trivially true.
So, the values are [tex]\( x = \frac{5}{6} \)[/tex] and [tex]\( y = -\frac{3}{4} \)[/tex].
### Part 3: Finding the Domain of the Function
Given the function [tex]\( f(x) = 4x + 7 \)[/tex] and the range [tex]\(\{311, 15, 19\}\)[/tex], we find the domain by solving each value in the range for [tex]\(x\)[/tex]:
1. For [tex]\(311\)[/tex]:
[tex]\[ 311 = 4x + 7 \Rightarrow 4x = 304 \Rightarrow x = 76 \][/tex]
2. For [tex]\(15\)[/tex]:
[tex]\[ 15 = 4x + 7 \Rightarrow 4x = 8 \Rightarrow x = 2 \][/tex]
3. For [tex]\(19\)[/tex]:
[tex]\[ 19 = 4x + 7 \Rightarrow 4x = 12 \Rightarrow x = 3 \][/tex]
So the domain is [tex]\(\{76, 2, 3\}\)[/tex].
### Part 4: Creating the Arrow Diagram
Given:
[tex]\[ M = \{e^8, 2\}, \quad N = \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
To find [tex]\( MM \times N \)[/tex]:
[tex]\[ M = \begin{pmatrix} e^8 \\ 2 \end{pmatrix} \][/tex]
[tex]\[ N = \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
We need to form the outer product:
[tex]\[ MM \times N = \begin{pmatrix} e^8 \\ 2 \end{pmatrix} \times \begin{pmatrix} a & b \\ y & 2 \end{pmatrix} \][/tex]
The resulting matrix is:
[tex]\[ \left[ \begin{array}{cc} e^8 \cdot a & e^8 \cdot b \\ 2 \cdot y & 2 \cdot 2 \end{array} \right] \][/tex]
[tex]\[ = \begin{pmatrix} e^8 a & e^8 b \\ 2y & 4 \end{pmatrix} \][/tex]
This is the outer product matrix.
In summary:
- Mean = 14
- Range = 21
- Values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]: [tex]\( x = \frac{5}{6}, y = -\frac{3}{4} \)[/tex]
- Domain = [tex]\(\{76, 2, 3\}\)[/tex]
- Arrow Diagram Result: [tex]\(\begin{pmatrix} e^8 a & e^8 b \\ 2y & 4 \end{pmatrix}\)[/tex]
