17. Write the SQL statement that will display the following (do not hardcode 'George', 'Lorraine', or any names, in your statement): | Person | Spouse | George | Lorraine
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- CUSTOMER Customer table is composed of customer number, name and phone number. Give and fill-in the appropriate attribute name for each of the column. custID 123 124 125 126 reservelD 5001 5002 5003 5004 5005 5006 5007 RESERVATION Each reservation is for one taxi. Reservation table is composed of reservation identification number, start reservation date, end reservation date, reservation days requested by customers starting from reservation date until end of reservation date, customer number that make the reservation and taxi number assigned to the reservation. Give and fill-in the appropriate attribute name for each of the column. taxill custName Ahmad Bin Abdullah Fatimah Binti Adam LAI LA2 Ruqayya Binti Idris Sulaiman Bin Daud LA3 LA4 startDate 01/10/2019 05/10/2019 05/10/2019 15/10/2019 20/10/2019 27/10/2019 02/11/2019 taxiType endDate 03/10/2019 12/10/2019 08/10/2019 17/10/2019 25/10/2019 Sedan Sedan Van Van 30/10/2019 04/11/2019 cust Phoneno TAXI Taxi table is composed of taxi…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?Update table product to set the value of column productType equal to the corresponding id value in the table productType. This requires a SELECT * FROM productType: to see what the id value is for each type. For example, in my database a. 1|meat b. 2| seafood c. 3| vegetable d. 4 grain and rice e. 5 bakery
- For the remaining questions, use the following table of people: people (id (pk), firstname, lastname, spouse_id (fk)) The table was created as: CREATE TABLE people (id INT NOT NULL AUTO INCREMENT PRIMARY KEY, firstname VARCHAR(30), lastname VARCHAR (30), spouse id INT, CONSTRAINT spouse fk FOREIGN KEY (spouse id) REFERENCES people (id)); This is some sample data: Participation Activity 9 +---- | id | firstname | lastname | spouse_id | + 1 | Marty 2 | Jennifer | McFly | Parker | McFly | Tannen | McFly | | 3 | Lorraine 4 | Biff 5 | George NULL | NULL | 5 | NULL | 3 | +----+--How many patients in the database have been diagnosed with acute bronchitis (ICD9Code = 466) and have a total of 3 or more diagnoses? (diagnosis table; COUNT(*), WHERE, GROUP BY, & HAVING clauses) / Result = 20List unique first (first_name) and last names (last_name) of customers (CUSTOMER table) who rented a movie (RENTAL table) between '2011-06-01' and '2012-01-01’ (rental_date). Sort by last_name. The common key is CUSTOMER_ID.
- Some rows of the STUDENT table are shown below: CODE NAME GPA YEAR 291 ALEX 3.1 2 938 MICHELE 2.3 1 931 JHON 3.3 1 182 JOE 3.4 2 190 REY 2.0 2 330 RON 3.9 3 Which best describes the result of the query below? SELECT YEAR, AVG(GPA) FROM STUDENT WHERE GPA > 2.0 GROUP BY YEAR a. The average maximum GPA of all the students in each year. b. The year of the student with the maximum GPA. c. The average GPA of students with a GPA higher than 2.0 each year. d. The average GPA of the students with a GPA higher than 2.0.DROP TABLE IF EXISTS Worker; CREATE TABLE Worker ( ); 0 INSERT ); WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST NAME TEXT, LAST NAME TEXT, SALARY INTEGER (15), JOINING DATE DATETIME, DEPARTMENT CHAR (25) INTO Worker (WORKER_ID, FIRST_NAME, LAST_NAME, SALARY, JOINING_DATE, DEPARTMENT) VALUES 'Arora', 100000, '14-02-20 09.00.00', 'HR'), (001, 'Monika', (002, 'Niharika', 'Verma' 80000, '14-06-11 09.00.00', 'Admin'), 'HR'), 'Admin'), Admin'), 'Account'), 'Kumar', 75000, '14-01-20 09.00.00', 'Account') DROP TABLE IF EXISTS Bonus; CREATE TABLE Bonus ( ); (003, 'Vishal' 'Singhal', 300000, '14-02-20 09.00.00', (004, 'Amitabh 'Singh', 500000, '14-02-20 09.00.00', (005, 'Vivek' 'Bhati', 500000, '14-06-11 09.00.00'. (006, 'Vipul', 'Diwan' 200000, '14-06-11 09.00.00', (007, 'Satish' (008, 'Geetika', 'Chauhan', 90000, '14-04-11 09.00.00', 'Admin'); WORKER_REF_ID INTEGER, BONUS AMOUNT INTEGER(10), BONUS_DATE DATETIME, FOREIGN KEY (WORKER_REF_ID) REFERENCES Worker (WORKER_ID) ON DELETE…The following table stores details of tutors, students and dates of lectures. The Primary Key is (TutorID, StudentID). TutorID TutorName StudentID StudentName Lecture_date T01 Kong S1052 Salim 10-May-19 T01 Kong S1051 Vikram 5-Jul-19 T02 Smith S1052 Salim 10-May-19 Explain why the above table is not in 2nd Normal Form. Normalize the table to 3rd Normal Form.
- Considering Vaccination Database, write queries to: Display the Subject CNIC, Name, Contact, Dose1 Center, and Dose2 Center as shown below: CNIC Name Contact Dose1 Center Dose2 Center 22401-6645321-1 Nasir 3409991112 Lachi Kohat 2 14301-6045321-5 Shahab 3409991112 kohat 3 Dara The subjects who have been vaccinated different types in Dose1 and Dose2. For instance, subjects who were vaccinated ‘Sinovac’ in the first dose, while Pfizer in the second dose.How many Viles of each type have been consumed so far. How many subjects are vaccinated from expired viles. Suggest the name of Incharge for the best performance award based on the highest number of subjects vaccinated on a single dayon this table CREATE TABLE students ( std_id int NOT NULL PRIMARY KEY, f_name VARCHAR2(20), l_name VARCHAR2(20), std_address VARCHAR2(50), tel_no VARCHAR2(15), SEX VARCHAR2(10), birthdate date ); use PL/SQL oracle to solve create A Function getAllData(arg): When Your call your function it should search about an item in your table and returns all data about it. arg is int all data are int and varchar2 and datein pl/sql ,use Cursor to get all the records from table job_history for employee_id 121