1. Write a query to find the number of employees Joined in each year? select to_char(hire_date,'YYYY'), COUNT(*) from employees group by to_char(hire_date,'YYYY'); 2. Write a query to display the number of people with the same job. SELECT job_id, COUNT(*) FROM employees GROUP BY job_id; 3. Display the highest, lowest, sum, and average salary of all employees. Label the columns Maximum, Minimum, Sum, and Average, respectively. Round your results to the nearest whole number. SELECT ROUND(MAX(salary),0) "Maximum", ROUND(MIN(salary),0) "Minimum", ROUND(SUM(salary),0) "Sum", ROUND(AVG(salary),0) "Average" FROM employees; 4. display the minimum, maximum, sum, and average salary for each job type. SELECT job_id, ROUND(MAX(salary),0) "Maximum", ROUND(MIN(salary),0) "Minimum", ROUND(SUM(salary),0) "Sum", ROUND(AVG(salary),0) "Average" FROM employees GROUP BY job_id; 5. Display the manager number and the salary of the lowest paid employee for that manager. Exclude anyone whose manager is not known. Exclude any groups where the minimum salary is $6,000 or less. Sort the output in descending order of salary. SELECT manager_id, MIN(salary) FROM employees WHERE manager_id IS NOT NULL GROUP BY manager_id HAVING MIN(salary) > 6000 ORDER BY MIN(salary) DESC; 6. Write a query that displays the difference between the highest and lowest salaries. Label the column DIFFERENCE. SELECT MAX(salary) - MIN(salary) DIFFERENCE FROM employees; 7. Determine the number of managers without listing them. Label the column Number of Managers. Hint: Use the MANAGER_ID column to determine the number of managers. SELECT COUNT(DISTINCT manager_id) "Number of Managers" FROM employees;