
Concept explainers
Grouping Rows:
SQL contains “GROUP BY” clause in order to group rows by common data. Though it is very powerful feature, it is hard to understand.
Syntax:
SELECT column_Name1 FROM table_Name GROUP BY column_Name2;
Example: Consider a table “student” contains two columns “student_Name” and “Department”. “GROUP BY” clause is used when there is a need to get the number of students from each department. The query for this scenario is given as follows.
SELECT department, COUNT (department) FROM student GROUP BY department;
When the above query is executed, number of students from each department will be displayed.
Aggregate Functions:
SQL has some built-in functions and they are called as aggregate functions. SQL contains five built-in functions. They are:
- SUM – This function is used to add values from the particular column.
- Syntax: SELECT SUM(column_Name) FROM table_Name;
- COUNT – This is used to count the number of rows for the particular column.
- Syntax: SELECT COUNT(column_Name) FROM table_Name;
- MAX – This function is used to get the maximum value from the column.
- Syntax: SELECT MAX(column_Name) FROM table_Name;
- MIN – This function is used to get the minimum value from the column.
- Syntax: SELECT MIN(column_Name) FROM table_Name;
- AVG – This function is used to get the average of all the values from the column.
- Syntax: SELECT AVG(column_Name) FROM table_Name;

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Chapter 7 Solutions
Database Systems: Design, Implementation, & Management
- What is business intelligence? Share the Business intelligence (BI) tools you have used and explain what types of decisions you made.arrow_forwardI need help fixing the minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place. My code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2 * pi * f; % Angular frequency % Parameters for the filters - let's adjust these to get more reasonable cutoffs R = 1e3; % Resistance in ohms (1 kΩ) C = 1e-6; % Capacitance in farads (1 μF) % For bandpass, we need appropriate L value for desired cutoffs L = 0.1; % Inductance in henries - adjusted for better bandpass response % Calculate cutoff frequencies first to verify they're in desired range f_cutoff_RC = 1 / (2 * pi * R * C); f_resonance = 1 / (2 * pi * sqrt(L * C)); Q_factor = (1/R) * sqrt(L/C); f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor)); f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor)); % Transfer functions % Low-pass filter (RC) H_low = 1 ./ (1 + 1i * w *…arrow_forwardTask 3. i) Compare your results from Tasks 1 and 2. j) Repeat Tasks 1 and 2 for 500 and 5,000 elements. k) Summarize run-time results in the following table: Time/size n String StringBuilder 50 500 5,000arrow_forward
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- Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781305627482Author:Carlos Coronel, Steven MorrisPublisher:Cengage Learning
