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Q57. Which of the following best describes the ON DELETE CASCADE referential integrity constraint? 

A. If a parent key is deleted, any child keys referenced by the parent key are automatically deleted. 

B. If any child key references a parent key, the record containing the parent key cannot be deleted. 

C. If a parent key isdeleted, all child keys are automatically set to a specified value. 

D. If a parent key is deleted, no test is made for referential integrity. 

Answer:


Q58. Consider the Orders relation shown in the exhibit. Which of the following SQL statements would return all complete tuples for order dates in 2002, arranged by amount from lowest to highest? 

A. SELECT * 

FROM Orders 

WHERE Order_Date LIKE _02 

ORDER BY Amount; 

B. SELECT (Order_Date, Amount) 

FROM Orders 

WHERE Order_Date LIKE %02 

ORDER BY Amount; 

C. SELECT * 

FROM Orders 

WHERE Order_Date LIKE _02 

ORDER BY Order_No; 

D. SELECT * 

FROM Orders 

WHERE Order_Date LIKE %02 

ORDER BY Amount; 

Answer:


Q59. Your enterprise has reached the conceptual design phase for a database project. What is the desired goal at the end of this design phase? 

A. A set of normalized relations 

B. A reviewed entity-relationship (ER) model 

C. An entity-relationship (ER) model with no redundant data 

D. A set ofdenormalized relations 

Answer:


Q60. Consider the Information Engineering diagram shown in the exhibit. Building_ID, R_ID, Room_Count and Room_Num are integer numbers, whereas Bldg_Name and Res_Name are represented by variable-length strings with a maximum of 20 characters. Location can be up to 50 characters long, and no building has more than 600 rooms. Which SQL statement best implements the BUILDING relation shown in this diagram? 

Building_ID NOT NULL PRIMARY KEY, Bldg_Name, Location, Room_Count); 

B. CREATETABLE BUILDING ( Building_ID NOT NULL PRIMARY KEY, Bldg_Name, Location, Room_Count, FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID)); 

C. CREATETABLE BUILDING (Building_ID INTEGER NOT NULL PRIMARY KEY, Bldg_Name VARCHAR (20), Location VARCHAR (50), 

D. CREATETABLE BUILDING (Building_ID INTEGER NOT NULL PRIMARY KEY, Bldg_Name VARCHAR (20), Location VARCHAR (50), 

Room_Count INTEGER CHECK (Room_Count > 

Room_Count INTEGER CHECK ( 

-1 And Room_Count < 601)); 

Room_Count > -1 Or Room_Count < 601)); 

Answer:


Q61. With regard to databases, what is normalization? 

A. The process of reducing the cardinality of a relation 

B. The process of organizing and refining relations 

C. The process of duplicating data to reduce the number of tables 

D. The process of limiting data stored in a table to a specific range of values 

Answer:


Q62. 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 

E. 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 

F. 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 

G. STUDENT( Student_Number: integer NOT NULL Name: variable length character string length 20 NOT NULL) Primary Key StudenMMumber 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 

Answer:


Q63. In which phase of database design do you identify entities, attribute domains, and relationships? 

A. Logical 

B. Physical 

C. Application 

D. Conceptual 

Answer:


Q64. 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 VIEWEmp_Dept AS SELECT Last_Name, First_Name, Dept_ID FROM Employee; 

B. UPDATE VIEWEmp_Dept AS SELECT * FROM Employee; 

C. UPDATE VIEWEmp_Dept AS SELECT Last_Name, First_Name, Dept_ID FROM Employee; 

D. CREATE VIEWEmp_Dept AS SELECT * FROM Employee WHERE ID = 0001 AND ID = 0002 AND ID = 0003 AND ID = 0004; 

Answer: