Get detailed and reliable answers to your questions with IDNLearn.com. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
