A.SELECT &1, "&2" FROM &3 WHERE last_name = '&4';
B.SELECT &1, '&2' FROM &3 WHERE '&last_name = '&4'';
C.SELECT &1, &2 FROM &3 WHERE last_name = '&4';
D.SELECT &1, '&2' FROM EMP WHERE last_name = '&4';

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.CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk FOREIGN KEY deptno REFERENCES dept deptno);
B.CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));
C.CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) NOT NULL, CONSTRAINT emp_deptno_fk REFERENCES dept (deptno) FOREIGN KEY (deptno));
D.CREATE TABLE EMP (empno NUMBER(4), ename VARCHAR2(35), deptno NUMBER(7,2) FOREIGN KEY CONSTRAINT emp_deptno_fk REFERENCES dept (deptno));

A.SELECT last_name, department_name FROM employees NATURAL JOIN departments;
B.SELECT last_name, department_name FROM employees JOIN departments ;
C.SELECT last_name, department_name FROM employees e JOIN departments d ON (e.department_id = d.department_id);
D.SELECT last_name, department_name FROM employees e RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id);
E.SELECT last_name, department_name FROM employees FULL JOIN departments ON (e.department_id = d.department_id);
F.SELECT last_name, department_name FROM employees e LEFT OUTER JOIN departments d ON (e.department_id = d.department_id);

A.SELECT ename, salary*12 'Annual Salary' FROM employees;
B.SELECT ename, salary*12 "Annual Salary" FROM employees;
C.SELECT ename, salary*12 AS Annual Salary FROM employees;
D.SELECT ename, salary*12 AS INITCAP("ANNUAL SALARY") FROM employees

A.Group functions on columns ignore NULL values.
B.Group functions on columns returning dates include NULL values.
C.Group functions on columns returning numbers include NULL values.
D.Group functions on columns cannot be accurately used on columns that contain NULL values.
E.Group functions on columns include NULL values in calculations if you use the keyword INC_NULLS.

A.You use a NEXTVAL pseudo column to look at the next possible value that would be generated from a sequence, without actually retrieving the value.
B.You use a CURRVAL pseudo column to look at the current value just generated from a sequence, without affecting the further values to be generated from the sequence.
C.You use a NEXTVAL pseudo column to obtain the next possible value from a sequence by actually retrieving the value from the sequence.
D.You use a CURRVAL pseudo column to generate a value from a sequence that would be used for a specified database column.
E.If a sequence starting from a value 100 and incremented by 1 is used by more than one application, then all of these applications could have a value of 105 assigned to their column whose value is being generated by the sequence.
F.You use a REUSE clause when creating a sequence to restart the sequence once it generates the maximum value defined for the sequence.

A.The SQL statement displays the desired results.
B.The column in the WHERE clause should be changed to display the desired results.
C.The operator in the WHERE clause should be changed to display the desired results.
D.The WHERE clause should be changed to use an outer join to display the desired results.

A.to find the tax percentage for each of the employees
B.to list the name, job_id, and manager name for all the employees
C.to find the name, salary, and the department name of employees who are not working with Smith
D.to find the number of employees working for the Administrative department and earning less than 4000
E.to display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned

A.SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco');
B.SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address;
C.SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address, customer_id;
D.SELECT city_address, COUNT(customer_id) FROM customers GROUP BY city_address IN ('Los Angeles', 'San Francisco');