Customers Customer ID Last Name C0001 Faulkner C0002 Wit Loan Id L0010 L0017 L0016 Customer Total L0019 10011 Customer Total Amount $200,000 $100,000 $150,000 $450,000 $10,000 $25,000 $35,000 The report above is grouped. Answer these questions:
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- C sharp Table: Student (the headers are the field names in the Students table) StudentID Name Age Gender ProgramID 791 Stephanie Brown 19 Female BCS 236 Shannon Dawn 25 Female BA 618 Geoff Berg 24 Male ARET 256 Andrew Schilling 22 Male BSC 902 Gary Sang 23 Male DAAD Note: There is a StudentDataSet with a Student table, a StudentTableAdapter, a StudentBindingSource, and a StudentDataGridView control on the form.Note: There is an Average query, named Average, that returns the average age of the student from the Student table.Note: There is also Max query, named Highest, that returns the highest age of the student from the Student table.Write the code you would place in the AverageButton click event on your form to call the Average query and the Highest query and display in a DifferenceLabel, the difference between the highest age of a student and the average age.The Driver Relationship team realized that maintaining driver IDs is difficult and requested an automatic way of incrementing the value when a new driver is added. You need to make the changes on the table to automatically increment the Driver ID. After the change, you need to insert the following driver: First Name: Nursin Last Name: Yilmaz Driving License ID: 4141447 Start Date: 2019-12-28 Driving License Checked: True Rating: 4.0MySql Workbench CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 1. List last name of all students whose first name is longer than 4 letters in ascending order accordingto the last name. Duplicated rows should be removed from the output.2. Count the total number of rooms in Location.3. Find the number of students in each major.4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...
- Given the tables: reading StationNo rainfall mintemp maxtemp 100 10 18 27 101 15 20 29 102 12 19 28 Stationregion station stationname 100 Savanne 101 Pamplemousses 102 Rose Belle (i) Write the statement to create the two table structures with the primary key as the field underlined. You may decide on the size of the fields. (ii) Write the statement to enter the data. (ii) Write the statement to display all the records with the fields station. stationname, mintemp, maxtemp .The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule. Hint: Perform a join on the Student and LessonSchedule tables, matching the student IDsThe database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule.
- Child row is a related row where there is a ? Primary key Foreign key Candidate key Composite keyTask 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Create a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the three fields. The CAR_ID and MAINTENACNE_TYPE_ID should be foreign keys to their original tables. Cascade update and cascade delete the foreign keys. Answer in MYSQL pleaseCreate table EMP and DEPT with the below mentioned structureStructure for EMP tableEmpIDNUMBER(4)DeptIDVARCHAR2(10)EmpNameCHAR(10)JobCHAR(10)HireDate DATESalary NUMBER(7, 2)Commission NUMBER (7, 2)Structure for DEPT tableDeptId VARCHAR2(10)Deptname VARCHAR2(20)No_of_Faculties NUMBER(2)In table EMP : EmpID should be the Primary Key and DeptID should be the foreign key.In table DEPT : DeptId should be the primary key. a) Find out the number of employees having “manager” as job.b) Display only the jobs with maximum salary greater than or equal to 3000c) Find all those employees whose job does not start with ‘M’. Find the names of the employees who are Managers and their date of joining is after “02-Jan-2006”.
- Having this information. StudentsAttribute Name Data TypeStudentID char(11)FirstName varchar(20)LastName varchar(20)Gender char(1)DateofBirth date CoursesAttribute Name Data TypeCourseCode varchar(6)CourseName varchar(70)Level char(2)Credits int RegistrationAttribute Name Data TypeStudentID char(11)CourseCode varchar(6)Grade decimal(2, 1) Write SQL code to: 1. How many courses are there on each level? 2. What is the average grade of courses that have been taken by the student with student ID 861103-2438? 3. Which students (studentID only) have the highest grade for the course ‘CS052’? 4. Find the courses (course codes only) that have been taken by both the student 861103-2438 and the student 123456-0980.Having this information. StudentsAttribute Name Data TypeStudentID char(11)FirstName varchar(20)LastName varchar(20)Gender char(1)DateofBirth date CoursesAttribute Name Data TypeCourseCode varchar(6)CourseName varchar(70)Level char(2)Credits int RegistrationAttribute Name Data TypeStudentID char(11)CourseCode varchar(6)Grade decimal(2, 1) Write SQL code to: 1. What are the student ID, the names, and birthday of the female students who were born before the last day of 1992? 2. How many students are registered in the course “PHY210”? 3. How many courses are offered by the department of Computer Science (i.e. course codes CS***)? 4. Which course(s) (couseCode and courseName) of level G1 give 5 credits?The Driver Relationship team realized that maintaining driver IDs is difficult and requested an automatic way of incrementing the value when a new driver is added. You need to make the changes on the table to automatically increment the Driver ID. After the change, you need to insert the following driver: First Name: Nursin Last Name: Yilmaz Driving License ID: 4141447 Start Date: 2019-12-28 Driving License Checked: True Rating: 4.0 THIS IS IN SQL FORMAT FOR CENGAGE MINDTAP, PLEASE HELP I AM STUCK
