Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You

Click the Exhibit button and examine the data in the EMPLOYEES and DEPARTMENTS tables.You want to retrieve all employees‘ last names, along with their managers‘ last names and their department names. Which query would you use?（）

A.SELECT last_name, manager_id, department_name FROM employees e FULL OUTER JOIN departments d ON (e.department_id = d.department_id);

B.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

C.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

D.SELECT e.last_name, m.last_name, department_name FROM employees e LEFT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);

E.SELECT e.last_name, m.last_name, department_name FROM employees e RIGHT OUTER JOIN employees m on ( e.manager_id = m.employee_id) RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);

F.SELECT last_name, manager_id, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id) ;

Click the Exhibit button and examine the data in the EMPLOYEES table.Examine the subquery:SELECT last_nameFROM employeesWHERE salary IN (SELECT MAX(salary)FROM employeesGROUP BY department_id);Which statement is true?（）

A.The SELECT statement is syntactically accurate.

B.The SELECT statement does not work because there is no HAVING clause.

C.The SELECT statement does not work because the column specified in the GROUP BY clause is not in the SELECT list.

D.The SELECT statement does not work because the GROUP BY clause should be in the main query and not in the subquery.

A.SELECT * FROM employees where salary > (SELECT MIN(salary) FROM employees GROUP BY department_id);

B.SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);

C.SELECT distinct department_id FROM employees WHERE salary > ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);

D.SELECT department_id FROM employees WHERE salary > ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);

E.SELECT last_name FROM employees WHERE salary > ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);

F.SELECT department_id FROM employees WHERE salary > ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));

A.SELECT * FROM employees where salary > (SELECT MIN(salary) FROM employees GROUP BY department_id);

B.SELECT * FROM employees WHERE salary = (SELECT AVG(salary) FROM employees GROUP BY department_id);

C.SELECT distinct department_id FROM employees WHERE salary > ANY (SELECT AVG(salary) FROM employees GROUP BY department_id);

D.SELECT department_id FROM employees WHERE salary > ALL (SELECT AVG(salary) FROM employees GROUP BY department_id);

E.SELECT last_name FROM employees WHERE salary > ANY (SELECT MAX(salary) FROM employees GROUP BY department_id);

F.SELECT department_id FROM employees WHERE salary > ALL (SELECT AVG(salary) FROM employees GROUP BY AVG(SALARY));

Click the Exhibit button and examine the data from the EMP table.The COMMISSION column shows the monthly commission earned by the employee.Which three tasks would require subqueries or joins in order to be performed in a single step?（）

A.deleting the records of employees who do not earn commission

B.increasing the commission of employee 3 by the average commission earned in department 20

C.finding the number of employees who do NOT earn commission and are working for department 20

D.inserting into the table a new employee 10 who works for department 20 and earns a commission that is equal to the commission earned by employee 3

E.creating a table called COMMISSION that has the same structure and data as the columns EMP_ID and COMMISSION of the EMP table

F.decreasing the commission by 150 for the employees who are working in department 30 and earning a commission of more than 800

Click the Exhibit button and examine the data from the EMP table.The COMMISSION column shows the monthly commission earned by the employee.Which two tasks would require subqueries or joins in order to be performed in a single step? （）

A.listing the employees who earn the same amount of commission as employee 3

B.finding the total commission earned by the employees in department 10

C.finding the number of employees who earn a commission that is higher than the average commission of the company

D.listing the departments whose average commission is more than 600

E.listing the employees who do not earn commission and who are working for department 20 in descending order of the employee ID

F.listing the employees whose annual commission is more than 6000

Click the Exhibit button and examine the data in the EMPLOYEES table.On the EMPLOYEES table, EMPLOYEE_ID is the primary key. MGR_ID is the ID of managers and refers to the EMPLOYEE_ID. The JOB_ID column is a NOT NULL column.Evaluate this DELETE statement:DELETE employee_id, salary, job_idFROM employeesWHERE dept_id = 90;Why does the DELETE statement fail when you execute it?（）

A.There is no row with dept_id 90 in the EMPLOYEES table.

B.You cannot delete the JOB_ID column because it is a NOT NULL column.

C.You cannot specify column names in the DELETE clause of the DELETE statement.

D.You cannot delete the EMPLOYEE_ID column because it is the primary key of the table.

Click the Exhibit button to examine the data of the EMPLOYEES table.Which statement lists the ID, name, and salary of the employee, and the ID and name of the employee‘s manager, for all the employees who have a manager and earn more than 4000?（）

A.SELECT employee_id "Emp_id", emp_name "Employee", salary, employee_id "Mgr_id", emp_name "Manager" FROM employees WHERE salary > 4000;

B.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m WHERE e.mgr_id = m.mgr_id AND e.salary > 4000;

C.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager" FROM employees e JOIN employees m ON (e.mgr_id = m.employee_id) AND e.salary > 4000;

D.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id", m.emp_name "Manager" FROM employees e SELF JOIN employees m WHERE e.mgr_id = m.employee_id AND e.salary > 4000;

E.SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.mgr_id "Mgr_id" m.emp_name "Manager" FROM employees e JOIN employees m USING (e.employee_id = m.employee_id) AND e.salary > 4000;

Click the Exhibit button to examine the data of the EMPLOYEES table. Evaluate this SQL statement:SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary, m.employee_id "Mgr_id", m.emp_name "Manager"FROM employees e JOIN employees m ON (e.mgr_id = m.employee_id)AND e.salary > 4000;What is its output?（）

A.A

B.B

C.C

D.D

E.E

Click the Exhibit button to examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.Two new departments are added to your company as shown:DEPARTMENT_ID DEPARTMENT_NAME MGR_ID LOCATION_ID9998 Engineering 1239999 Administrative BostonYou need to list the names of employees, the department IDs, the department names, and the cities where the departments are, even if there are no employees in the departments and even if the departments are not yet assigned to a location. You need to join the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables to retrieve this information.Which statement do you execute to retrieve this information?（）

A.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d RIGHT OUTER JOIN employees e ON d.department_id = e.department_id RIGHT OUTER JOIN locations l ON d.location_id = l.location_id;

B.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d FULL OUTER JOIN employees e ON d.department_id = e.department_id FULL OUTER JOIN locations l ON d.location_id = l.location_id;

C.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d LEFT OUTER JOIN employees e ON d.department_id = e.department_id LEFT OUTER JOIN locations l ON d.location_id = l.location_id;

D.SELECT last_name, department_id, department_name, city FROM departments d NATURAL JOIN employees e NATURAL JOIN locations l;