OCP CBT Question 498

Please solve the following question.

You are querying the `EMPLOYEES` table. You want to display 'High Salary' for employees with a salary greater than 10000, 'Medium Salary' for salaries between 5000 and 10000 (inclusive), and 'Low Salary' for salaries less than 5000. Which SQL statement correctly achieves this using a `CASE` expression?

EMPLOYEES Table Structure`sqlDESC EMPLOYEES;Name Null? Type----------------- -------- -------------EMPLOYEE_ID NOT NULL NUMBER(6)SALARY NUMBER(8,2)...`

SELECT employee_id, salary, CASE salary WHEN > 10000 THEN 'High Salary' WHEN BETWEEN 5000 AND 10000 THEN 'Medium Salary' ELSE 'Low Salary' END AS salary_category FROM employees;

SELECT employee_id, salary, CASE WHEN salary > 10000 THEN 'High Salary' WHEN salary >= 5000 AND salary <= 10000 THEN 'Medium Salary' ELSE 'Low Salary' END AS salary_category FROM employees;

SELECT employee_id, salary, CASE salary WHEN salary > 10000 THEN 'High Salary' WHEN salary >= 5000 THEN 'Medium Salary' ELSE 'Low Salary' END AS salary_category FROM employees;

SELECT employee_id, salary, DECODE(salary, > 10000, 'High Salary', BETWEEN 5000 AND 10000, 'Medium Salary', 'Low Salary') AS salary_category FROM employees;

EMPLOYEES Table StructuresqlDESC EMPLOYEES;Name Null? Type----------------- -------- -------------EMPLOYEE_ID NOT NULL NUMBER(6)SALARY NUMBER(8,2)...

EMPLOYEES Table Structure`sqlDESC EMPLOYEES;Name Null? Type----------------- -------- -------------EMPLOYEE_ID NOT NULL NUMBER(6)SALARY NUMBER(8,2)...`