Product Description
Exam Number/Code: 1D0-541
Exam name: CIW v5 Database Design Specialist
1D0-541 test is the important of CIW certifaction. Only you pass the 1D0-541 exam you could have the chance to get CIW certifaction. To help examinee, Exambible publish the Questions and Answers about 1D0-541 exam. And the 83 simulation exam are all designed by expert from Exambible. Examinees could have an enough prepare by these simulation exercises and pass the 1D0-541 test successfully.
Exambible also provide free 1D0-541 Demo, examinees can download and use before buying 1D0-541 Q&As, then decide if they need to purchase.
More CIW CIW Certification Resources
Recommended CIW exams
- CIW 1D0-475 exam: MASTER-CIW ADMINISTRATOR FOR MISROSOFT AND NOVELL PROFESSNLS
- CIW 1D0-525 exam: CIW v5 E-Commerce Designer
- CIW 1D0-532 exam: CIW WEB DEVELOPER(JCERT)
- CIW 1D0-410 exam: CIW Foundations
- CIW 1D0-437 exam: CIW PERL FUNDAMENTALS
- CIW 1D0-460 exam: CIW INTERNETWORKING PROFESSIONAL
- CIW 1D0-510 exam: CIW v5 Founfations Exam
- CIW 1D0-420 exam: CIW SITE DESIGNER
- CIW 1D0-470 exam: Security Analyst CIW SECURITY PROFESSIONAL
- CIW 1D0-450 exam: CIW SERVER ADMINISTRATOR
- CIW 1D0-435 exam: CIW JAVA XCRIPT FUNDAMENTALS
- CIW 1D0-441 exam: CIW DATABASE SPECIALIST
CIW CIW Certification 1D0-541 Web Demo
This webdemo is just a demo data, only for reference and learning, there is no other purposes.
1.What is the highest normal form of the relation(s) shown in the exhibit?
A. Second normal form
B. First normal form
C. Boyce-Codd normal form
D. Third normal form
Answer: A
3. Consider the entity-relation (ER) diagram shown in the exhibit. When
the logical database design phase is completed, which of the following
is a valid DBDL description of the base relations for the ER diagram?
A. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
B. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
Foreign Key Class_Num References STUDENT
C. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
STU_CLASS(
Student_Number: integer NOT NULL
Class_Num: integer NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
D. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
STU_CLASS(
Student_Number: integer NOT NULL
Class_Num: integer NOT NULL)
Primary Key Student_Number, Class_Num
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
Answer: D
4. Consider the relations shown in the exhibit. Which of the following
SQL statements would enter data from the Customers relation into the
Atlanta_Customers relation?
A. INSERT INTO Atlanta_Customers
VALUES(
SELECT *
FROM Customer s
WHERE Sales_Office = tlanta
B. INSERT INTO Atlanta_Customers
SELECT *
FROM Customers
WHERE Sales_Office = tlanta
C. INSERT INTO Atlanta_Customers
SELECT Cust_No, Cust_Name, Satisfaction_Rate, Sales_Rep_No
FROM Customers
WHERE Sales_Office = tlanta
D. INSERT INTO Atlanta_Customers
SELECT Cust_No, Cust_Name, Sales_Office, Sales_Rep_No
FROM Customers
WHERE Sales_Office = tlanta
Answer: C
5. Consider the relation shown in the exhibit. Which of the following
SQL statements would properly remove all tuples for New York customers?
A. DELETE *
FROM Customers
WHERE Sales_Office = New York;
B. DELETE
FROM Customers
WHERE Sales_Office = ew York?WHERE Sales_Office = ?ew York?
C. DELETE *
FROM Customers
WHERE Sales_Office = ew York?WHERE Sales_Office = ?ew York?
D. DELETE
FROM Customers
WHERE Sales_Office NOT LIKE ew York? WHERE Sales_Office NOT LIKE ?ew York?
Answer: B
6. Consider the Information Engineering diagram shown in the exhibit. Which DBDL definition best describes this diagram?
A. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key Building_ID
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key R_ID
B. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key BUILDING
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key RESIDENT
C. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key BUILDING
Foreign Key BUILDING(Building_ID) references RESIDENT(Building_ID)
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key RESIDENT
D. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key Building_ID
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key R_ID
Foreign Key Building_ID references BUILDING(Building_ID)
Answer: D
7. Several SQL operations are performed by User 1 to access the Fee
information for Bowling in the Act_Fee relation (shown in the exhibit).
The first access returns a fee of 50. An unrelated SQL operation by
another user updates the Bowling fee to 60. The second access by User 1
returns a fee of 60. What problem has occurred?
A. Rollback
B. Deadlock
C. Dirty read
D. No problem has occurred.
Answer: D
8. Consider the Employee relation shown in the exhibit. A database
manager wants to set up a view called Emp_Dept that allows users to
find employees and their department ID numbers. Which SQL statement
will accomplish this?
A. CREATE VIEW Emp_Dept
AS SELECT Last_Name, First_Name, Dept_ID
FROM Employee;
B. UPDATE VIEW Emp_Dept
AS SELECT *
FROM Employee;
C. UPDATE VIEW Emp_Dept
AS SELECT Last_Name, First_Name, Dept_ID
FROM Employee;
D. CREATE VIEW Emp_Dept
AS SELECT *
FROM Employee
WHERE ID = 0001
AND ID = 0002
AND ID = 0003
AND ID = 0004;
Answer: A
2. Which three pieces of information did E.F. Codd describe as necessary to retrieve a data value from a relational database?
A. Attribute, domain, and tuple
B. Entity, relation name, and domain
C. Table name, primary key, and entity
D. Attribute, relation name, and primary key
Answer: D
