(a)
To find the relationship between students, courses and faculty members in the specified relation.
(a)

Explanation of Solution
The given relation is
The above relation Student and Faculty has many-to-many relationship since a new table is generated whose primary key is the combination of all primary key.
Since a faculty teaches many students and a student is taught by many faculties. Therefore, the relation between student and faculty is many-to-many relationship.
The relation between Student and Course has many to many relationships since a student can take many courses and a course can be taken by many students.
After creating new table, the collection of tables is as follows:
(b)
To find the relationship between students, courses and faculty members in the specified relation.
(b)

Explanation of Solution
The given relation is as follows:
Since one student can take many courses and one course is taken by many students.
Therefore, the attributes StudentNum and CourseNum in the Student relation used as the primary key shows the many-to-many relationship between StudentFaculty and Course tables.
The relation Faculty and Course has many-to-many relationship since one faculty can be assigned many course numbers and one course number can be assigned to many faculties.
After creating new table, the collection of tables is as follows:
(c)
To find the relationship between students, courses and faculty members in the specified relation.
(c)

Explanation of Solution
The given relation is as follows:
The relation StudentFaculty and Faculty has many-to-many relationship since many students have assigned many courses.
The relation Faculty and Course has many-to-many relationship as many faculties can be assigned many courses.
The relation StudentFaculty and Course has many-to-many relationship since one faculty can be assigned many course numbers and one course number can be assigned to many faculties.
Also, the relation StudentFaculty and Faculty has many to many relationships since one student is taught by many faculties and one faculty can be assigned to many students.
(d)
To find the relationship between students, courses and faculty members in the specified relation.
(d)

Explanation of Solution
The given relation is
The above relation Student and Faculty has many-to-many relationship since a new table is generated whose primary key is the combination of all primary key.
Since a faculty teaches many students and a student is taught by many faculties. Therefore, the relation between student and faculty is many-to-many relationship.
The relation between Student and Course has many tomany relationships since a student can take many courses and a course can be taken by many students.
After creating new table the collection of tables are as follows:
(e)
To find the relationship between students, courses and faculty members in the specified relation.
(e)

Explanation of Solution
The given relation is as follows:
The relation Student and Course has many-to-many relationship since many students have assigned many courses.
The relation Faculty and Course has many-to-many relationship since many faculties can be assigned to many subjects.
The relation Student and Faculty has many-to-many relationship since one faculty can be assigned many students and one student can be assigned to many faculties.
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Chapter 6 Solutions
CONCEPTS OF DATABASE MANAGEMENT
- Implement the code In MATLAB and send a picture of the Implementation from within the program Simulate data from magnetic sensor % magnetic FieldStrength = 0.5; % Magnetic field strength in Tesla Classify materials based on magnetic % field strength if magnetic FieldStrength > 0.3 material = 'Metal'; % Detect metal (e.g., iron) else material = 'Non-metal'; % Non-metal materials end ; disp(['Detected material: ', material])arrow_forwardImplement the code In MATLAB and send a picture of the Implementation from within the program Simulate infrared absorbance values % IR Absorbance = 0.75; % Infrared absorbance of the material Classify material based on infrared % absorbance if IR Absorbance > 0.7 material = 'Plastic'; % Plastic absorbs more IR material = 'Other'; % Other materials else like wood or metal end ;disp(['Material detected: ', material])arrow_forwardImplement the code In MATLAB and send a picture of the Implementation from within the program MATLAB code to detect magnetic materials % Assume we have a reading from a magnetic % sensor magnetic field = 0.8; % Magnetic field strength in Tesla If the material is magnetic (like iron), % there will be a higher reading if magnetic field > 0.5 'material = 'Magnetic (Metal) material = 'Non-Magnetic (Plastic/ else ; 'Wood) end ;disp(['The material is: ', material])arrow_forward
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