1. Write a query to display the last name and hire date of any employee in the same
   department as Zlotkey. Exclude Zlotkey.
	SELECT last_name, hire_date
	FROM employees
	WHERE department_id = (SELECT department_id
	FROM employees
	WHERE last_name = ’Zlotkey’)
	AND last_name <> ’Zlotkey’;
2. Create a query to display the employee numbers and last names of all employees who earn more
   than the average salary. Sort the results in ascending order of salary.

	SELECT employee_id, last_name
	FROM employees
	WHERE salary > (SELECT AVG(salary)
			FROM employees)
	ORDER BY salary;

3. Write a query that displays the employee numbers and last names of all employees who work in a
   department with any employee whose last name contains a u. Place your SQL statement in a text
   file named lab6_3.sql. Run your query.

	SELECT employee_id, last_name
	FROM employees
	WHERE department_id IN (SELECT department_id
				FROM employees
				WHERE last_name like ’%u%’);

4. Display the last name, department number, and job ID of all employees whose department location
   ID is 1700.

	SELECT last_name, department_id, job_id
	FROM employees
	WHERE department_id IN (SELECT department_id
				FROM departments
				WHERE location_id = 1700);

5. Display the last name and salary of every employee who reports to King.

	SELECT last_name, salary
	FROM employees
	WHERE manager_id = (SELECT employee_id
			    FROM employees
			    WHERE last_name = ’King’);

6. Display the department number, last name, and job ID for every employee in the Executive
   department.

	SELECT department_id, last_name, job_id
	FROM employees
	WHERE department_id IN (SELECT department_id
				FROM departments
				WHERE department_name = ’Executive’);

7. Modify the query in lab6_3.sql to display the employee numbers, last names, and salaries of
   all employees who earn more than the average salary and who work in a department with any
   employee with a u in their name. Resave lab6_3.sql to lab6_7.sql. Run the statement in
   lab6_7.sql.

	SELECT employee_id, last_name, salary
	FROM employees
	WHERE department_id IN (SELECT department_id
				FROM employees
				WHERE last_name like ’%u%’)
	AND salary > (SELECT AVG(salary)FROM employees);