Choose the letter of the correct answer based on the table EMPLOYEES as shown be
ID: 3902307 • Letter: C
Question
Choose the letter of the correct answer based on the table EMPLOYEES as shown below.
Table 1.0 EMPLOYEES
1.
Which of the following is the correct query that display the Job_id and add the ff: function:
Count the total number of job_id per distinct values; Compute for the summary of salary per job_id; and Compute for the average salary per job_id
Select one:
a. SELECT COUNT (DISTINCT(JOB_ID) AS “NO. OF JOB_ID”, SUM(SALARY) AS “TOTAL SALARY”, AVG(SALARY) AS “AVERAGE SALARY” FROM EMPLOYEES GROUP BY JOB_ID;
b. SELECT DISTINCT(JOB_ID), COUNT * (JOB_ID) AS “NO. OF JOB_ID”, SUM(SALARY) AS “TOTAL SALARY”, AVG(SALARY) AS “AVERAGE SALARY” FROM EMPLOYEES GROUP BY JOB_ID;
c. SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS “NO. OF JOB_ID”, SUM(SALARY) AS “TOTAL SALARY”, AVG(SALARY) AS “AVERAGE SALARY” FROM EMPLOYEES GROUP BY JOB_ID;
d. SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS “NO. OF JOB_ID”, SUM(SALARY) AS “TOTAL SALARY”, AVG(SALARY) AS “AVERAGE SALARY” FROM EMPLOYEES;
2.
Which of the following is the correct that display distinct job_id and the the total number per distinct (unique) job_id.
Select one:
a. SELECT DISTICT(JOB_ID), COUNT(JOB_ID) AS “TOTAL NUMBER OF JOB_ID”
FROM EMPLOYEES;
b. SELECT DISTICT((JOB_ID), COUNT(JOB_ID)) AS “TOTAL NUMBER OF JOB_ID”
FROM EMPLOYEES
GROUP BY JOB_ID;
c. SELECT DISTICT(JOB_ID), COUNT*(JOB_ID) AS “TOTAL NUMBER OF JOB_ID”
FROM EMPLOYEES
GROUP BY JOB_ID;
d. SELECT DISTICT(JOB_ID), COUNT(JOB_ID) AS “TOTAL NUMBER OF JOB_ID”
FROM EMPLOYEES
GROUP BY JOB_ID;
3.
Which of the following is the correct query that displays the MINIMUM salary of employees per job_id? Note job_id should be in lowercase.
Select one:
a. SELECT MIN(SALARY) AS LOWEST SALARY, LOWER(JOB_ID)
FROM EMPLOYEES
GROUP BY JOB_ID;
b. SELECT MIN(SALARY) AS “LOWEST SALARY”, LOWER(JOB_ID)
FROM EMPLOYEES
GROUP BY JOB_ID;
c. SELECT MIN(SALARY) AS “LOWEST SALARY”, LOWER(JOB_ID)
FROM EMPLOYEES;
d. SELECT MIN(SALARY) AS LOWEST SALARY, LOWER(JOB_ID)
FROM EMPLOYEES;
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1. Answer is c.
SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS “NO. OF JOB_ID”, SUM(SALARY) AS “TOTAL SALARY”, AVG(SALARY) AS “AVERAGE SALARY”
FROM EMPLOYEES
GROUP BY JOB_ID;
Distinct job ID's will be selected and the respective count, sum and average for the same will be generated, by grouping the JOB_ID's.
2. Answer is d.
SELECT DISTINCT(JOB_ID), COUNT(JOB_ID) AS “TOTAL NUMBER OF JOB_ID”
FROM EMPLOYEES
GROUP BY JOB_ID;
displays distinct job_id and the the total number per distinct (unique) job_id.
3. Answer is b.
SELECT MIN(SALARY) AS “LOWEST SALARY”, LOWER(JOB_ID)
FROM EMPLOYEES
GROUP BY JOB_ID;
displays the MINIMUM salary of employees per job_id, where job_id will be displayed in lower case.