Product Description
Exam Number/Code: 000-512
Exam name: db2 udb v7.1 family fundamentals
000-512 test is the important of IBM certification. Only you pass the 000-512 exam you could have the chance to get IBM certification. To help examinee, Exambible publish the Questions and Answers about 000-512 exam. And the 110 simulation exam are all designed by expert from Exambible. Examinees could have an enough prepare by these simulation exercises and pass the 000-512 test successfully.
Exambible also provide free 000-512 Demo, examinees can download and use before buying 000-512 Q&As, then decide if they need to purchase.
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IBM Solutions Expert 000-512 Web Demo
This webdemo is just a demo data, only for reference and learning, there is no other purposes.
1.Given the following SQL statements:
CREATE TABLE tab1 (col1 INT)
CREATE TABLE tab2 (col1 INT)
INSERT INTO tab1 VALUES (NULL),(1)
INSERT INTO tab2 VALUES (NULL),(1)
SELECT COUNT(*) FROM tab1
WHERE col1 IN
(SELECT col1 FROM tab2)
Which of the following is the result of the SELECT COUNT(*) statement?
A. 1
B. 2
C. 3
D. 4
E. 0
Answer: A
3. Given a table T1, with a column C1 char(3), that contains strings in
upper and lower case letters, which of the following queries will find
all rows where C1 is the string 'ABC' in any case?
A. SELECT * FROM t1 WHERE c1 = 'ABC'
B. SELECT * FROM t1 WHERE UCASE(c1) = 'ABC'
C. SELECT * FROM t1 WHERE IGNORE_CASE(c1 = 'ABC')
D. SELECT * FROM t1 WHERE c1 = 'ABC' WITH OPTION CASE INSENSITIVE
Answer: B
4. Which of the following occurs if an application ends abnormally during an active unit of work?
A. Current unit of work is committed
B. Current unit of work is rolled back
C. Current unit of work remains active
D. Current unit of work moves to pending state
Answer: B
5. User2 has DBADM authority on database DB1. This allows the user to do which of the following?
A. Drop database DB1
B. Backup database DB1
C. Create tables in any database
D. Create tables in database DB1
Answer: D
6. Given the tables:
COUNTRY
ID NAME PERSON CITIES
1 Argentina 1 10
2 Canada 2 20
3 Cuba 2 10
4 Germany 1 0
5 France 7 5
STAFF
ID LASTNAME
1 Jones
2 Smith
The statement:
SELECT * FROM staff, country
will return how many rows?
A. 2
B. 4
C. 5
D. 7
E. 10
Answer: E
7. Given the following table definition:
STAFF
id INTEGER
name CHAR(20)
dept INTEGER
job CHAR(20)
years INTEGER
salary DECIMAL(10,2)
comm DECIMAL(10,2)
The job column contains these job types: manager, clerk, and
salesperson. Which of the following statements will return the data
with all managers together, all clerks together and all salespeople
together in the output?
A. SELECT * FROM staff ORDER BY job
B. SELECT job, name FROM staff GROUP BY name, job
C. SELECT * FROM staff GROUP BY name, job, id, dept, years, salary, comm
D. SELECT * FROM staff ORDER BY name, job, id, dept, years, salary, comm
Answer: A
8. Given the two following tables:
Names
Name Number
Wayne Gretzky 99
Jaromir Jagr 68
Bobby Orr 4
Bobby Hull 23
Brett Hull 16
Mario Lemieux 66
Steve Yzerman 19
Claude Lemieux 19
Mark Messier 11
Mats Sundin 13
Points
Name Points
Wayne Gretzky 244
Jaromir Jagr 68
Bobby Orr 129
Bobby Hull 93
Brett Hull 121
Mario Lemieux 189
Joe Sakic 94
Which of the following statements will display the player Names,
numbers and points for all players with an entry in both tables? Which
of the following statements will display the player? Names, numbers and
points for all players with an entry in both tables?
A. SELECT names.name, names.number, points.points FROM names INNER JOIN points ON names.name=points.name
B. SELECT names.name, names.number, points.points FROM names FULL OUTER JOIN points ON names.name=points.name
C. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
D. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
E. SELECT names.name, names.number, points.points FROM names FULL OUTER JOIN points ON names.name=points.name
F. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
G. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
H. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
I. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
J. SELECT names.name, names.number, points.points FROM names LEFT OUTER JOIN points ON names.name=points.name
K. SELECT names.name, names.number, points.points FROM names RIGHT OUTER JOIN points ON names.name=points.name
Answer: A
2. Which of the following describes why savepoints are NOT allowed inside an atomic unit of work?
A. Atomic units of work span multiple databases, but savepoints are limited to units of work which operate on a single database.
B. A savepoint implies that a subset of the work may be allowed to
succeed, while atomic operations must succeed or fail as a unit.
C. A savepoint requires an explicit commit to be released, and commit
statements are not allowed in atomic operations such as compound SQL.
D. A savepoint cannot be created without an active connection to a
database, but atomic operations can contain a CONNECT as a
sub-statement.
Answer: B
