1)d One result from many rows per group.
2)c. Reduce disk I/O for SELECT statements.
3)a. CREATE SEQUENCE patient_id_seq
START WITH 1000
MAXVALUE
4)d Users with access.
5)c SELECT a.divisionno “division”,
(a.num_work/COUNT(*))* 100 “%workers”,
(a.sal_sum/COUNT(*))*100 “%salary”.
FROM
(SELECT divisionno,COUNT(*)num_work,SUM(SAL)sal_sum
FROM scott.work
GROUP BY divisionno)a;
6) b. The worker table column corresponding to the region table contains null values for rows that need to be displayed
7) b.Use a sub query in the WHERE clause to compare the salary value.
8) b. A multiple-row sub query has been used with a single row comparison operator.
9) a Describe work
10) d. A syntax error results.
2)c. Reduce disk I/O for SELECT statements.
3)a. CREATE SEQUENCE patient_id_seq
START WITH 1000
MAXVALUE
4)d Users with access.
5)c SELECT a.divisionno “division”,
(a.num_work/COUNT(*))* 100 “%workers”,
(a.sal_sum/COUNT(*))*100 “%salary”.
FROM
(SELECT divisionno,COUNT(*)num_work,SUM(SAL)sal_sum
FROM scott.work
GROUP BY divisionno)a;
6) b. The worker table column corresponding to the region table contains null values for rows that need to be displayed
7) b.Use a sub query in the WHERE clause to compare the salary value.
8) b. A multiple-row sub query has been used with a single row comparison operator.
9) a Describe work
10) d. A syntax error results.
