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- You have the following tables: APARTMENTS (ADDRESS, CITY, STATE, RENTER_ID, RENTER_LAST_NAME) RENTERS (RENTER_ID, FIRST_NAME, LAST_NAME) What is the primary key for the APARTMENTS table? (It may be a composite key involving 2 or more fields) What are the foreign keys, if any? What is the primary key for the RENTERS table? What are the foreign keys, if any? What field in RENTERS can be eliminated so that the RENTERS table is normalized?Problem 22 Insert the following customer into the CUST_MYSQL table, allowing the AUTO_INCREMENT attribute set up in Problem 20.a and Problem 20.b to generate the customer number automatically: CUST_LNAME CUST_FNAME CUST_BALANCE Powers Ruth 500Convert the following table to an equivalent collection of tables that are in third normal form. This table contains information about patients of a dentist. Each patient belongs to a household. Patient (HouseholdNum, HouseholdName, Street, City, State, PostalCode, Balance, PatientNum, PatientName, (ServiceCode, Description, Fee, Date)) The following dependencies exist in the Patient table: PatientNum → HouseholdNum, HouseholdName, Street, City, State, PostalCode, Balance, PatientName HouseholdNum → HouseholdName, Street, City, State, PostalCode, Balance ServiceCode → Description, Fee PatientNum, ServiceCode → Date
- Problem 22 Insert the following customer into the CUST_MYSQL table, allowing the AUTO_INCREMENT attribute set up in Problem 20.a and Problem 20.b to generate the customer number automatically: CUST_LNAME CUST_FNAME CUST_BALANCE Powers Ruth 500 This is what problem 20.a said, and got right: Using MySQL, create a table named CUST_MYSQL with the same fields as in Problem 16, except, use the AUTO_INCREMENT feature for the CUST_NUM field. This is what problem 20.b said, and got right: Using MySQL, alter the table to populate the CUST_NUM field, beginning the increment with 2000.Write SQL query to create the.given table:Worker(Worker Id int primary key, Name varchar(15), Department varchar(10))Write SQL query to create the.given table: Worker(Worker Id int primary key, Name varchar(15), Department varchar(10))
- Some rows of the table STUDENT are shown below: CODE NAME DEPARTMENT GPA YEAR 291 ALEX PSI 3.1 1992 938 MICHELE PHY 2.3 1992 931 JHON MD 3.3 2001 182 JOE MD 3.4 2002 190 REY PHY 2.0 2001 330 RON PSI 3.9 2001 You want the name all departments that have a department GPA average of 2004 greater than the global GPA average. Which query gives you the desired result? a. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT); b. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 AND AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT) GROUP BY DEPARTMENT; c. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT GPA FROM STUDENT GROUP BY DEPARTMENT); d. SELECT * FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT);Given the relation members(mid: INTEGER, name: CHAR(30), age: INTEGER, tier: INTEGER) Write a SQL query to show the average age in each tier with more than 2 members, including how many members in each tier, for tiers above 5 in ascending order, with labels for aggregate fields shown.Write an SQL query to fetch “FIRST_NAME” from Worker table using the alias name as <WORKER_NAME>. ii.Write an SQL query to fetch unique values of DEPARTMENT from Worker table.
- You have two tables teachers and students. Each student belongs to a teacher. Complete the query to join the tables on the teacher id. SELECT * FROM students JOIN teachers ON _________ ; a. Students.teacher_id = teachers.id b. teachers_id = id c. teachers.id = student.id d. teachers = students.idThis is the problem: (from Database Systems Book, 13th ed) Write a query to display the author last name, first name, book title, and replacement cost for each book. Sort the results by book number and then author ID (Figure P7.92). (25 rows)Write MS SQL query to retrieve exployee with second highest salary from employee table Employee(Id, name, salary)