Database Systems Design Implementation And Management 12Th Edition by Carlos Coronel -Test Bank A+

Description

1. Normalization works through a series of stages called normal forms.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. Normalization is a process that is used for changing attributes to entities.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. In order to meet performance requirements, portions of the database design may need to be occasionallydenormalized.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. Denormalization produces a lower normal form.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. Normalization is a very important database design ingredient, and the highest level is always the most desirable.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Moderate REF: p.202

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Database Tables and Normalization

1. Reporting anomalies in a table can cause a multitude of problems for managers and can be fixed through applicationprogramming.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.205

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Need For Normalization

1. Data redundancy produces data anomalies.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.206

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Need For Normalization

1. The objective of normalization is to ensure that each table conforms to the concept of well-formed relations.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.206

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Relational models view data as part of a table or collection of tables in which all key values must be identified.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.208

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Repeating groups must be eliminated by ensuring that each row defines a single entity.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.208

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A dependency of one nonprime attribute on another nonprime attribute is a partial dependency.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Dependency diagrams are very helpful in getting a bird’s­eye view of all the relationships among a table’s attributes.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Dependencies that are based on only a part of a composite primary key are called transitive dependencies.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. All relational tables satisfy the 1NF requirements.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.211

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. In the context of partial dependencies, data redundancies occur because every row entry requires duplication ofdata.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.211

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Since a partial dependency can exist only if a table’s primary key is composed of several attributes, if a table in 1NFhas a single-attribute primary key, then the table is automatically in 2NF.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Moderate REF: p.212

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. It is possible for a table in 2NF to exhibit transitive dependency, where the primary key may rely on one or morenonprime attributes to functionally determine other nonprime attributes.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Moderate REF: p.212

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. A determinant is any attribute whose value determines other values within a column.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.213

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Data stored at their highest level of granularity are said to be atomic data.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. Atomic attributes are attributes that can be further subdivided.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. A table is in BCNF if every determinant in the table is a foreign key.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. A table is in fourth normal form if it is in third normal form and has no independent multivalued dependencies.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. Normalization represents a micro view of the entities within the ERD.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Normalization and Database Design

1. The combination of normalization and ER modeling yields a useful ERD, whose entities can be translated intoappropriate relationship structures.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.229

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Normalization and Database Design

1. A good relational DBMS excels at managing denormalized relations.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.229

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. The advantage of higher processing speed must be carefully weighed against the disadvantage of data anomalies.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.229

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. Normalization purity is often easy to sustain in the modern database environment.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.231

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. Unnormalized database tables often lead to various data redundancy disasters in production databases.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.232

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. Attributes should clearly define participation, connectivity, and document cardinality.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.233

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data-Modeling Checklist

1. Normalization works through a series of stages called normal forms. For most purposes in business databasedesign, stages are as high as you need to go in the normalization process.
   1. two three
   2. four d. five

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. From a structural point of view, 3NF is better than .
   1. 4NF 2NF
   2. 5NF d. 6NF

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. From a structural point of view, 2NF is better than .
   1. 1NF 3NF
   2. 4NF d. BCNF

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. An attribute that is part of a key is known as a(n)
   1. important nonprime
   2. prime d. entity

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. A table that displays data redundancies yields .
   1. consistencies anomalies
   2. fewer attributes d. more entities

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.205

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Need For Normalization

1. Data redundancy produces .
   1. slower lookups robust design
   2. efficient storage use d. data integrity problems

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.206

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Need For Normalization

1. Attribute A attribute B if all of the rows in the table that agree in value for attribute A also agree in valuefor attribute B.
   1. determines derives from
   2. controls d. owns

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.207

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. Some very specialized applications may require normalization beyond the .
   1. 1NF 2NF
   2. 3NF d. 4NF

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.207

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Of the following normal forms, is mostly of theoretical interest.
   1. 1NF 3NF
   2. BCNF d. DKNF

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.207

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A table that has all key attributes defined, has no repeating groups, and all its attributes are dependent on theprimary key is said to be in .
   1. 1NF 2NF
   2. 3NF d. 4NF

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.207

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A(n) exists when there are functional dependencies such that XY is functionally dependent on WZ, X isfunctionally dependent on W, and XY is the primary key.
   1. atomic attribute repeating group
   2. partial dependency d. transitive dependency

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.207

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A(n) exists when there are functional dependencies such that Y is functionally dependent on X, Z isfunctionally dependent on Y, and X is the primary key.
   1. partial dependency repeating group
   2. atomic attribute d. transitive dependency

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.208

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A derives its name from the fact that a collection of multiple entries of the same type can exist for anysingle key attribute occurrence.
   1. partial dependency transitive dependency
   2. repeating group d. primary key

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.208

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A relational table must not contain a(n) .
   1. entity attribute
   2. relationship d. repeating group

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.208

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. In a(n) diagram, the arrows above the attributes indicate all desirable dependencies.
   1. Chen dependency
   2. functionality d. ER

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Dependencies based on only a part of a composite primary key are known as
   1. primary partial
   2. incomplete d. composite

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.211

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. If a table has multiple candidate keys and one of those candidate keys is a composite key, the table can have

based on this composite candidate key even when the primary key chosen is a single attribute.

1. Boyce-Codd normal forms redundancies
2. time-variances d. partial dependencies

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.215

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A table that is in 2NF and contains no transitive dependencies is said to be in .
   1. 1NF 2NF
   2. 3NF d. 4NF

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.215

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Improving leads to more flexible queries.
   1. atomicity normalization
   2. denormalization d. derived attribute

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. An atomic attribute .
   1. cannot exist in a relational table cannot be further subdivided
   2. displays multiplicity d. is always chosen to be a foreign key

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. The most likely data type for a surrogate key is .
   1. character date
   2. logical d. numeric

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. Granularity refers to .
   1. the size of a tableb. the level of detail represented by the values in a table’srow
2. the number of rows in atabled. the number of attributes represented in a table

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving the Design

1. From a system functionality point of view, attribute values can be calculated when they are needed to writereports or invoices.
   1. derived atomic
   2. granular d. historical

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.217

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving The Design

1. In a real-world environment, we must strike a balance between design integrity and .
   1. robustness flexibility
   2. uniqueness d. ease of use

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.220

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Surrogate Key Considerations

1. For most business transactional databases, we should normalize relations into .
   1. 1NF 2NF
   2. 3NF d. 6NF

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.220

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. To generate a surrogate key, Microsoft Access uses a(n) data type.
   1. character sequence
   2. AutoNumber d. identity

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.220

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Surrogate Key Considerations

1. A table where every determinant is a candidate key is said to be in .
   1. BCNF 2NF
   2. 1NF d. 4NF

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.221

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. BCNF can be violated only if the table contains more than one
   1. primary candidate
   2. foreign d. secondary

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.221

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. When a table contains only one candidate key, are considered to be equivalent.
   1. the 1NF and the 2NF the 3NF and the BCNF
   2. the 4NF and the 3NF d. the BCNF and the DKNF

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.221

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. In a situation, one key determines multiple values of two other attributes and those attributes areindependent of each other.
   1. multivalued dependency transitive dependency
   2. partial dependency d. functional dependency

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.225

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. A table where all attributes are dependent on the primary key but are independent of each other, and no rowcontains two or more multivalued facts about an entity is said to be in .
   1. 1NF 2NF
   2. 3NF d. 4NF

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. A table is in 4NF if it is in 3NF, and .
   1. all attributes must be dependent on the primary key and must be dependent on each other
   2. all attributes are unrelated
   3. it has no multivalued dependencies
   4. no column contains the same values

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. When designing a database, you should .
   1. make sure that entities are in normal form before table structures are created
   2. create table structures then normalize the database
   3. only normalize the database when performance problems occur
   4. consider more important issues such as performance before normalizing

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.226

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Normalization and Database Design

1. An example of denormalization is using a denormalized table to hold report data. This is required whencreating a tabular report in which the columns represent data that are stored in the table as rows.
   1. transitive 3NF
   2. component d. temporary

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.231

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. The conflicts between design efficiency, information requirements, and performance are often resolved through_____.
   1. compromises that include normalization conversion from 2NF to 3NF
   2. compromises that include denormalization d. conversion from 3NF to 4NF

ANSWER: c

PTS: 1 DIF: Difficulty: Moderate REF: p.231

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Denormalization

1. Data warehouse routinely uses structures in its complex, multilevel, multisource data environment.
   1. 1NF 2NF
   2. 3NF d. 4NF

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.232

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. databases reflect the ever-growing demand for greater scope and depth in the data on which decisionsupport systems increasingly rely.
   1. Normalized Data warehouse
   2. Temporary d. Report

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.232

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. If database tables are treated as though they were files in a file system, the never has a chance todemonstrate its superior data-handling capabilities.

ANSWER: RDBMS

relational database management system

relational database management system (RDBMS)RDBMS (relational database management system)

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database and Normalization

1. The price paid for increased performance through denormalization is a larger amount of .

ANSWER: redundancy

data redundancy

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. In order to meet requirements, you may have to denormalize some portions of a database design.

ANSWER: performance

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. is a process to help reduce the likelihood of data anomalies.

ANSWER: Normalization

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. Any attribute that is at least part of a key is known as a (n)_____ .

ANSWER: prime attribute

key attribute

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. When designing a new database structure based on the business requirements of the end users, the databasedesigner will construct a data model using a technique such as .

ANSWER: Crow’s Foot notation ERDs

PTS: 1 DIF: Difficulty: Easy REF: p.202

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Database Tables and Normalization

1. The is central to a discussion of normalization.

ANSWER: concept of keys

PTS: 1 DIF: Difficulty: Easy REF: p.206

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. A dependency based on only a part of a composite primary key is called a (n)_____ .

ANSWER: partial dependency

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. The problem with transitive dependencies is that they still yield data .

ANSWER: anomalies

PTS: 1 DIF: Difficulty: Easy REF: p.210

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. All relational tables satisfy the

ANSWER: 1NF

first normal form

first normal form (1NF)

1NF (first normal form)

PTS: 1 DIF: Difficulty: Easy REF: p.211

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Because a partial dependency can exist only when a table’s primary key is composed of several attributes, a tablewhose key consists of only a single attribute is automatically in 2NF once it is in 1NF.

ANSWER: primary

PTS: 1 DIF: Difficulty: Easy REF: p.212

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. Any attribute whose value determines other values within a row is known as a .

ANSWER: determinant

PTS: 1 DIF: Difficulty: Easy REF: p.213

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: The Normalization Process

1. An attribute that cannot be further subdivided is said to display .

ANSWER: atomicity

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving The Design

1. refers to the level of detail represented by the values stored in a table’s row.

ANSWER: Granularity

PTS: 1 DIF: Difficulty: Easy REF: p.216

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving The Design

1. In a real-world environment, changing granularity requirements might dictate changes in primary key selection, andthose changes might ultimately require the use of __

ANSWER: surrogate

PTS: 1 DIF: Difficulty: Easy REF: p.217

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Improving The Design

1. It becomes difficult to create a suitable key when the related table uses a composite primary key.

ANSWER: foreign

PTS: 1 DIF: Difficulty: Easy REF: p.219

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Surrogate Key Considerations

1. When a nonkey attribute is the determinant of a key attribute, the table is in 3NF but not in .

ANSWER: BCNF

Boyce-Codd normal form

Boyce-Codd normal form

(BCNF)BCNF (Boyce-Codd normal form)

PTS: 1 DIF: Difficulty: Easy REF: p.221

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Surrogate Key Considerations

1. In the , no row may contain two or more multivalued facts about an entity.

ANSWER: 4NF

fourth normal form

fourth normal form (4NF)

4NF (fourth normal form)

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Higher-Level Normal Forms

1. An ERD is created through a (n)________

ANSWER: iterative

PTS: 1 DIF: Difficulty: Easy REF: p.226

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Normalization and Database Design

1. The combination of and ER modeling yields a useful ERD, whose entities may now be translated intoappropriate table structures.

ANSWER: normalization

PTS: 1 DIF: Difficulty: Easy REF: p.229

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Normalization and Database Design

1. Unnormalized tables yield no simple strategies for creating virtual tables known as .

ANSWER: views

PTS: 1 DIF: Difficulty: Easy REF: p.232

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Denormalization

1. According to the data-modeling checklist, should be nouns that are familiar to business, should be short andmeaningful, and should document abbreviations, synonyms, and aliases for each entity.

ANSWER: entity names

PTS: 1 DIF: Difficulty: Easy REF: p.233

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data-Modeling Checklist

1. Explain normalization and its different forms.

ANSWER: Normalization is a process for evaluating and correcting table structures to minimize data redundancies,thereby reducing the likelihood of data anomalies. The normalization process involves assigningattributes to tables based on the concept of determination. Normalization works through a series ofstages called normal forms. The first three stages are described as first normal form (1NF), secondnormal form (2NF), and third normal form (3NF). From a structural point of view, 2NF is better than1NF, and 3NF is better than 2NF. For most purposes in business database design, 3NF is as high as youneed to go in the normalization process. However, you will discover that properly designed 3NFstructures also meet the requirements of fourth normal form (4NF).

PTS: 1 DIF: Difficulty: Moderate REF: p.202

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Database Tables and Normalization

1. What characteristics do tables that conform to the concept of well-informed relations have?

ANSWER: Tables that conform to the concept of well-informed relations have the following characteristics:

1. Each table represents a single subject.
2. No data item will be unnecessarily stored in more than one table. This results in tables that havelower redundancies. The reason for this requirement is to ensure that the data is updates in onlyone place.
3. All nonprime attributes in a table are dependent on the primary key alone. The reason for thisrequirement is to ensure that the data is uniquely identifiable by a primary key value.
4. Each table is void of insertion, update, or deletion anomalies, which ensure the integrity andconsistency of the data.

PTS: 1 DIF: Difficulty: Moderate REF: p.206

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. Describe a dependency diagram and explain its purpose.

ANSWER: Dependency diagrams are very helpful in getting a bird’s eye view of all the relationships among atable’s attributes, and their use makes it less likely that you will overlook an important dependency.The following are features of a dependency diagram:

1. The primary key attributes are bold, underlined, and shaded in a different color.
2. The arrows above the attributes indicate all desirable dependencies—that is, dependencies basedon the primary key.
3. The arrows below the dependency diagram indicate less desirable dependencies. Two types ofsuch dependencies exist:
   1. Partial dependencies. A dependency based on only a part of a composite primary key is apartial dependency.
   2. Transitive dependencies. A transitive dependency is a dependency of one nonprime attributeon another nonprime attribute. The problem with transitive dependencies is that they still yielddata anomalies.

PTS: 1 DIF: Difficulty: Moderate REF: p.210

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. What steps are involved in the conversion to third normal form?

ANSWER: Step 1: Make New Tables to Eliminate Transitive Dependencies

For every transitive dependency, write a copy of its determinant as a primary key for a new table. Adeterminant is any attribute whose value determines other values within a row. If you have threedifferent transitive dependencies, you will have three different determinants. As with the conversion to2NF, it is important for the determinant remain in the original table to serve as a foreign key.

Step 2: Reassign Corresponding Dependent Attributes

Identify the attributes that are dependent on each determinant identified in Step 1. Place the dependentattributes in the new tables with their determinants and remove them from their original tables.

PTS: 1 DIF: Difficulty: Moderate REF: p.213

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: The Normalization Process

1. Explain the Boyce-Codd normal form (BCNF). How is it related to other normal forms?

ANSWER: A table is in Boyce-Codd normal form (BCNF) when every determinant in the table is a candidate key.A candidate key has the same characteristics as a primary key, but for some reason, it was not chosento be the primary key. Clearly, when a table contains only one candidate key, the 3NF and the BCNFare equivalent. In other words, BCNF can be violated only when the table contains more than onecandidate key. Most designers consider the BCNF to be a special case of the 3NF. In fact, if thetechniques shown in this chapter are used, most tables conform to the BCNF requirements once the3NF is reached.

PTS: 1 DIF: Difficulty: Moderate REF: p.221

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Higher-Level Normal Forms

1. Explain how database designers design and normalize databases.

ANSWER: First, an ERD is created through an iterative process. Database designers begin by identifying relevantentities, their attributes, and their relationships. Then they use the results to identify additional entitiesand attributes. The ERD provides the big picture, or macro view, of an organization’s data requirementsand operations. Second, normalization focuses on the characteristics of specific entities; that is,normalization represents a micro view of the entities within the ERD. Also, the normalization processmight yield additional entities and attributes to be incorporated into the ERD. Therefore, it is difficult toseparate normalization from ER modeling; the two techniques are used in an iterative and incrementalprocess.

©2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website or school-approved learning management system for classroom use.

CHAPTER 7: INTRODUCTION TO STRUCTURED QUERY LANGUAGE (SQL)

1. A database language enables the user to perform complex queries designed to transform the raw data into usefulinformation.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.247

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. SQL is considered difficult to learn; its command set has a vocabulary of more than 300 words.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.247

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. The ANSI prescribes a standard SQL–the current fully approved version is known as SQL­07.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.248

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. The ANSI SQL standards are also accepted by the ISO.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.248

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. Data type selection is usually dictated by the nature of the data and by the intended use.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.252

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. Only numeric data types can be added and subtracted in SQL.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.252

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. Entity integrity is enforced automatically when the primary key is specified in the CREATE TABLE commandsequence.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.258

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. The CHECK constraint is used to define a condition for the values that the attribute domain cannot have.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.260

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. You cannot insert a row containing a null attribute value using SQL.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.264

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. SQL requires the use of the ADD command to enter data into a table.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.264

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. Any changes made to the contents of a table are not physically saved on disk until you use the SAVE <table name>command.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.265

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. To list the contents of a table, you must use the DISPLAY command.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.266

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. The COMMIT command does not permanently save all changes. In order to do that, you must use SAVE.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.266

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. All SQL commands must be issued on a single line.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.267

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. Although SQL commands can be grouped together on a single line, complex command sequences are best shown on separate lines, with space between the SQL command and the command’s components.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Moderate REF: p.267

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Command

1. If you have not yet used the COMMIT command to store the changes permanently in the database, you canrestore the database to its previous condition with the ROLLBACK command.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.269

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. You can select partial table contents by naming the desired fields and by placing restrictions on the rows to beincluded in the output.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.271

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Oracle users can use the Access QBE (query by example) query generator.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.271

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Mathematical operators cannot be used to place restrictions on character-based attributes.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.273

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. String comparisons are made from left to right.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.274

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Date procedures are often more software-specific than other SQL procedures.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.274

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. SQL allows the use of logical restrictions on its inquiries such as OR, AND, and NOT.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.277

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. ANSI-standard SQL allows the use of special operators in conjunction with the WHERE clause.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.279

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. The conditional LIKE must be used in conjunction with wildcard characters.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.280

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Most SQL implementations yield case-insensitive searches.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.281

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Some RDBMSs, such as Microsoft Access, automatically make the necessary conversions to eliminate casesensitivity.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.281

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. The COUNT function is designed to tally the number of non-null “values” of an attribute, and is often used inconjunction with the DISTINCT clause.
   1. True
   2. False

ANSWER: True

PTS: 1 DIF: Difficulty: Easy REF: p.293

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional select Query Keywords

1. An alias cannot be used when a table is required to be joined to itself in a recursive query.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Moderate REF: p.303

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Joining Database Tables

1. When joining three or more tables, you need to specify a join condition for one pair of tables.
   1. True
   2. False

ANSWER: False

PTS: 1 DIF: Difficulty: Easy REF: p.303

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Joining Database Tables

1. The SQL data manipulation command HAVING:
   1. restricts the selection of rows based on a conditional expression.
   2. restricts the selection of grouped rows based on a condition.
   3. modifies an attribute’s values in one or more table’s rows.
   4. groups the selected rows based on one or more attributes.

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.248

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. The SQL command that allows a user to permanently save data changes is .
   1. INSERT SELECT
   2. COMMIT d. UPDATE

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.248

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. The constraint assigns a value to an attribute when a new row is added to a table.
   1. CHECK UNIQUE
   2. NOT NULL d. DEFAULT

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.260

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. The constraint assigns a value to an attribute when a new row is added to a table.
   1. CHECK UNIQUE
   2. NOT NULL d. DEFAULT

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.260

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. The SQL command that allows a user to list the contents of a table is .
   1. INSERT SELECT
   2. COMMIT d. UPDATE

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.266

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. In Oracle, the command is used to change the display for a column, for example, to place a $ in front of anumeric value.
   1. DISPLAY FORMAT
   2. CHAR d. CONVERT

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.267

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. UPDATE tablename

*****

[WHERE conditionlist];

The command replaces the ***** in the syntax of the UPDATE command, shown above.

1. SET columnname = expression columnname = expression
2. expression = columnname d. LET columnname = expression

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.268

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. An example of a command a user would use when making changes to a PRODUCT table is .
   1. CHANGE PRODUCT

SET P_INDATE = ‘18-JAN-2004’WHERE P_CODE = ‘13-Q2/P2’;

1. ROLLBACK PRODUCT

SET P_INDATE = ‘18-JAN-2004’WHERE P_CODE = ‘13-Q2/P2’;

1. EDIT PRODUCT

SET P_INDATE = ‘18-JAN-2004’WHERE P_CODE = ‘13-Q2/P2’;

1. UPDATE PRODUCT

SET P_INDATE = ‘18-JAN-2004’WHERE P_CODE = ‘13-Q2/P2’;

ANSWER: d

PTS: 1 DIF: Difficulty: Moderate REF: p.268

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. The command is used to restore the database to its previous condition.
   1. COMMIT; RESTORE; COMMIT; BACKUP;
   2. COMMIT; ROLLBACK; d. ROLLBACK;

ANSWER: d

PTS: 1 DIF: Difficulty: Easy REF: p.269

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. Some RDBMSs, such as Oracle, automatically data changes when issuing data definition commands.
   1. COMMIT ROLLBACK
   2. UNSAVE d. UPDATE

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.269

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. When a user issues the DELETE FROM tablename command without specifying a WHERE condition, .
   1. no rows will be deleted the first row will be deleted
   2. the last row will be deleted d. all rows will be deleted

ANSWER: d

PTS: 1 DIF: Difficulty: Moderate REF: p.270

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. The command would be used to delete the table row where the P_CODE is ‘BRT-345’.
   1. DELETE FROM PRODUCT

WHERE P_CODE = ‘BRT-345’;

1. REMOVE FROM PRODUCTWHERE P_CODE = ‘BRT-345’;
2. ERASE FROM PRODUCT

WHERE P_CODE = ‘BRT-345’;

1. ROLLBACK FROM PRODUCTWHERE P_CODE = ‘BRT-345’;

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.270

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. A(n) is a query that is embedded inside another query.
   1. alias operator
   2. nested d. view

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.270

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. Which of the following queries will output the table contents when the value of V_CODE is equal to 21344?
   1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE <> 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE <= 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE => 21344;

ANSWER: c

PTS: 1 DIF: Difficulty: Moderate REF: p.271

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. Which of the following is used to select partial table contents?
   1. SELECT <column(s)>

FROM <Table name>

BY <Conditions>;

1. LIST <column(s)>

FROM <Table name>

BY <Conditions>;

1. SELECT <column(s)>

FROM <Table name>

WHERE <Conditions>;

1. LIST<column(s)>

FROM <Table name>

WHERE <Conditions>;

ANSWER: c

PTS: 1 DIF: Difficulty: Moderate REF: p.271

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. Which of the following queries will output the table contents when the value of V_CODE is not equal to 21344?
   1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE <> 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE <= 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE => 21344;

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.271

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. Which of the following queries will output the table contents when the value of the character field P_CODE

isalphabetically less than 1558-QW1?

1. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICEFROM PRODUCT

WHERE P_CODE <‘1558-QW1’;

1. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICEFROM PRODUCT

WHERE P_CODE = [1558-QW1];

1. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICEFROM PRODUCT

WHERE P_CODE = (1558-QW1);

1. SELECT P_CODE, P_DESCRIPT, P_QOH, P_MIN, P_PRICEFROM PRODUCT

WHERE P_CODE = {1558-QW1};

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.273

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. Which of the following queries will list all the rows in which the inventory stock dates occur on or after January 20,2016?
   1. SELECT P_DESCRIPT, P_QOH, P_MIN, P_PRICE, P_INDATEFROM PRODUCT

WHERE P_INDATE >= ‘20-JAN-2016’;

1. SELECT P_DESCRIPT, P_QOH, P_MIN, P_PRICE, P_INDATEFROM PRODUCT

WHERE P_INDATE >= $20-JAN-2010$;

1. SELECT P_DESCRIPT, P_QOH, P_MIN, P_PRICE, P_INDATEFROM PRODUCT

WHERE P_INDATE <= ‘20-JAN-2010’;

1. SELECT P_DESCRIPT, P_QOH, P_MIN, P_PRICE, P_INDATEFROM PRODUCT

WHERE P_INDATE >= {20-JAN-2010};

ANSWER: a

PTS: 1 DIF: Difficulty: Moderate REF: p.274

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. Which of the following queries will use the given columns and column aliases from the PRODUCT table todetermine the total value of inventory held on hand?
   1. SELECT P_DESCRIPT, P_QOH, P_PRICE, P_QOH/P_PRICEFROM PRODUCT;
   2. SELECT P_DESCRIPT, P_QOH, P_PRICE, P_QOH=P_PRICEFROM PRODUCT;
   3. SELECT P_DESCRIPT, P_QOH, P_PRICE, P_QOH*P_PRICEFROM PRODUCT;
   4. SELECT P_DESCRIPT, P_QOH, P_PRICE, P_QOH-P_PRICEFROM PRODUCT;

ANSWER: c

PTS: 1 DIF: Difficulty: Moderate REF: p.275

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. A(n) is an alternate name given to a column or table in any SQL statement.
   1. alias data type
   2. stored function d. trigger

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.275

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. Which of the following queries uses the correct SQL syntax to list the table contents for either V_CODE = 21344or V_CODE = 24288?
   1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344OR V_CODE <= 24288;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344OR V_CODE => 24288;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344OR V_CODE > 24288;

1. SELECT P_DESCRIPT, P_INDATE, P_PRICE, V_CODEFROM PRODUCT

WHERE V_CODE = 21344OR V_CODE = 24288;

ANSWER: d

PTS: 1 DIF: Difficulty: Moderate REF: p.277

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Select Queries

1. According to the rules of precedence, which of the following computations should be completed first?
   1. performing additions and subtractions
   2. performing multiplications and divisions
   3. performing operations within parentheses
   4. performing power operations

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.277

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. The special operator used to check whether an attribute value is within a range of values is .
   1. BETWEEN NULL
   2. LIKE d. IN

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.279

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. The special operator used to check whether an attribute value matches a given string pattern is .
   1. BETWEEN IS NULL
   2. LIKE d. IN

ANSWER: c

PTS: 1 DIF: Difficulty: Easy REF: p.279

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. The special operator used to check whether a subquery returns any rows is .
   1. BETWEEN EXISTS
   2. LIKE d. IN

ANSWER: b

PTS: 1 DIF: Difficulty: Easy REF: p.279

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. All changes in a table structure are made using the command, followed by a keyword that produces thespecific changes a user wants to make.
   1. ALTER TABLE UPDATE TABLE
   2. COOMIT TABLE d. ROLLBACK TABLE

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.283

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. A table can be deleted from the database by using the
   1. DROP TABLE DELETE TABLE
   2. MODIFY TABLE d. ERASE TABLE

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.290

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Data Definition Commands

1. The SQL query to output the contents of the EMPLOYEE table sorted by last name, first name, and initial is

_____.

1. SELECT EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_AREACODE, EMP_PHONEFROM EMPLOYEE

LIST BY EMP_LNAME, EMP_FNAME, EMP_INITIAL;

1. SELECT EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_AREACODE, EMP_PHONEFROM EMPLOYEE

ORDER BY EMP_LNAME, EMP_FNAME, EMP_INITIAL;

1. SELECT EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_AREACODE, EMP_PHONEFROM EMPLOYEE

DISPLAY BY EMP_LNAME, EMP_FNAME, EMP_INITIAL;

1. SELECT EMP_LNAME, EMP_FNAME, EMP_INITIAL, EMP_AREACODE, EMP_PHONEFROM EMPLOYEE

SEQUENCE BY EMP_LNAME, EMP_FNAME, EMP_INITIAL;

ANSWER: b

PTS: 1 DIF: Difficulty: Moderate REF: p.291

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Additional Select Query Keywords

1. Which of the following queries is used to list a unique value for V_CODE, where the list will produce only a list ofthose values that are different from one another?
   1. SELECT ONLY V_CODE

FROM PRODUCT;

1. SELECT UNIQUE V_CODEFROM PRODUCT;
2. SELECT DIFFERENT V_CODEFROM PRODUCT;
3. SELECT DISTINCT V_CODEFROM PRODUCT;

ANSWER: d

PTS: 1 DIF: Difficulty: Moderate REF: p.292

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Additional Select Query Keywords

1. The SQL aggregate function that gives the number of rows containing non-null values for a given column is

.

1. COUNT MIN
2. MAX d. SUM

ANSWER: a

PTS: 1 DIF: Difficulty: Easy REF: p.293

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Select Query Keyword

1. The query to join the P_DESCRIPT and P_PRICE fields from the PRODUCT table and the V_NAME,V_AREACODE, V_PHONE, and V_CONTACT fields from the VENDOR table where the values of V_CODEmatch is .
   1. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONEFROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE <> VENDOR.V_CODE;

1. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONEFROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE = VENDOR.V_CODE;

1. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONEFROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE <= VENDOR.V_CODE;

1. SELECT P_DESCRIPT, P_PRICE, V_NAME, V_CONTACT, V_AREACODE, V_PHONEFROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE => VENDOR.V_CODE;

ANSWER: b

PTS: 1 DIF: Difficulty: Moderate REF: p.301

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Additional Select Query Keyword

1. The query to join the P_DESCRIPT and P_PRICE fields from the PRODUCT table and the V_NAME,V_AREACODE, V_PHONE and V_CONTACT fields from the VENDOR table, where the values of V_CODEmatch and the output is ordered by the price is .
   1. SELECT P_DESCRIPT, PRODUCT.P_PRICE, VENDOR.V_NAME,VENDOR.V_CONTACT, VENDOR.V_AREACODE, VENDOR.V_PHONE

FROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE <> VENDOR.V_CODE;ORDER BY PRODUCT.P_PRICE;

1. SELECT P_DESCRIPT, PRODUCT.P_PRICE, VENDOR.V_NAME,VENDOR.V_CONTACT, VENDOR.V_AREACODE, VENDOR.V_PHONE

FROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE => VENDOR.V_CODE;ORDER BY PRODUCT.P_PRICE;

1. SELECT P_DESCRIPT, PRODUCT.P_PRICE, VENDOR.V_NAME,VENDOR.V_CONTACT, VENDOR.V_AREACODE, VENDOR.V_PHONE

FROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE <= VENDOR.V_CODE;ORDER BY PRODUCT.P_PRICE;

1. SELECT P_DESCRIPT, PRODUCT.P_PRICE, VENDOR.V_NAME,VENDOR.V_CONTACT,

VENDOR.V_AREACODE, VENDOR.V_PHONEFROM PRODUCT, VENDOR

WHERE PRODUCT.V_CODE = VENDOR.V_CODE;ORDER BY PRODUCT.P_PRICE;

ANSWER: d

PTS: 1 DIF: Difficulty: Moderate REF: p.302

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Additional Select Query Keyword

1. The basic SQL vocabulary has fewer than

ANSWER: 100

one hundred

a hundred

PTS: 1 DIF: Difficulty: Easy REF: p.247

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. In the SQL environment, the word covers both questions and actions.

ANSWER: query

PTS: 1 DIF: Difficulty: Easy REF: p.248

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Introduction to SQL

1. A(n) is a logical group of database objects, such as tables and indexes, that are related to each other.

ANSWER: schema

PTS: 1 DIF: Difficulty: Easy REF: p.251

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. With the exception of the database process, most RDBMS vendors use SQL that deviates little from theANSI standard SQL.

ANSWER: creation

creating

PTS: 1 DIF: Difficulty: Easy REF: p.251

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. ___ is the process the DBMS uses to verify that only registered users access the database.

ANSWER: Authentication

PTS: 1 DIF: Difficulty: Easy REF: p.251

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. S. state abbreviations are always two characters, so (2) is a logical choice for the data type representing astate column.

ANSWER: CHAR

PTS: 1 DIF: Difficulty: Easy REF: p.252

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. The SQL data type DATE stores date in the date format.

ANSWER: Julian

PTS: 1 DIF: Difficulty: Easy REF: p.254

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. To make SQL code more _______, most SQL programmers use one line per column (attribute) definition.

ANSWER: readable

PTS: 1 DIF: Difficulty: Easy REF: p.255

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. In a 1:M relationship, a user must always create the table for the side first.

ANSWER: 1

one

PTS: 1 DIF: Difficulty: Easy REF: p.256

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. The specification is used to avoid having duplicated values in a column.

ANSWER: UNIQUE

PTS: 1 DIF: Difficulty: Easy REF: p.257

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. __ words are words used by SQL to perform specific functions.

ANSWER: Reserved

PTS: 1 DIF: Difficulty: Easy REF: p.258

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. A common practice is to create a(n) on any field that is used as a search key, in comparison operations in aconditional expression, or when a user wants to list rows in a specific order.

ANSWER: index

PTS: 1 DIF: Difficulty: Easy REF: p.263

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. To delete an index, one must use the

ANSWER: DROP INDEX

PTS: 1 DIF: Difficulty: Easy REF: p.264

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Definition Commands

1. In an INSERT command, a user can indicate just the attributes that have required values by listing the insideparentheses after the table name.

ANSWER: attribute names

names

PTS: 1 DIF: Difficulty: Easy REF: p.265

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. A(n) character is a symbol that can be used as a general substitute for other characters or commands.

ANSWER: wildcard

wild card

PTS: 1 DIF: Difficulty: Easy REF: p.266

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. A(n) , also known as a nested query or an inner query, is a query that is embedded (or nested) inside anotherquery.

ANSWER: subquery

PTS: 1 DIF: Difficulty: Easy REF: p.270

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Data Manipulation Commands

1. The command, coupled with appropriate search conditions, is an incredibly powerful tool that enables a userto transform data into information.

ANSWER: SELECT

PTS: 1 DIF: Difficulty: Easy REF: p.271

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. DATE() and SYSDATE are special functions that return today’s date in MS Access and , respectively.

ANSWER: Oracle

PTS: 1 DIF: Difficulty: Easy REF: p.276

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. In SQL, all expressions evaluate to true or false.

ANSWER: conditional

PTS: 1 DIF: Difficulty: Easy REF: p.278

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. A specialty field in mathematics, known as algebra, is dedicated to the use of logical operators.

ANSWER: Boolean

PTS: 1 DIF: Difficulty: Easy REF: p.278

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. If a user adds a new column to a table that already has rows, the existing rows will default to a value of forthe new column.

ANSWER: null

PTS: 1 DIF: Difficulty: Easy REF: p.285

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Data Definition Commands

1. A table can be deleted from the database by using the

ANSWER: DROP TABLE

PTS: 1 DIF: Difficulty: Easy REF: p.290

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Data Definition Commands

1. A(n) order sequence is a multilevel ordered sequence that can be created easily by listing several attributes,separated by commas, after the ORDER BY clause.

ANSWER: cascading

PTS: 1 DIF: Difficulty: Easy REF: p.291

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Select Query Keywords

1. Rows can be grouped into smaller collections quickly and easily using the clause within the SELECTstatement.

ANSWER: GROUP BY

PTS: 1 DIF: Difficulty: Easy REF: p.297

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Select Query Keywords

1. The clause of the GROUP BY statement operates very much like the WHERE clause in the SELECTstatement.

ANSWER: HAVING

PTS: 1 DIF: Difficulty: Easy REF: p.298

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Additional Select Query Keywords

1. A(n) is performed when data are retrieved from more than one table at a time.

ANSWER: join

PTS: 1 DIF: Difficulty: Easy REF: p.300-301

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Joining Database Tables

1. An alias is especially useful when a table must be joined to itself in a(n)

ANSWER: recursive

PTS: 1 DIF: Difficulty: Easy REF: p.303

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Joining Database Tables

1. The condition is generally composed of an equality comparison between the foreign key and the primary keyof related tables.

ANSWER: join

PTS: 1 DIF: Difficulty: Moderate REF: p.301

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Joining Database Tables

1. Explain the two SQL functions.

ANSWER: 1. SQL is a data definition language (DDL). It includes commands to create database objects such astables, indexes, and views, as well as commands to define access rights to those databases objects.

1. SQL is a data manipulation language (DML). It includes commands to insert, update, delete, andretrieve data within the database tables.

PTS: 1 DIF: Difficulty: Moderate REF: p.247

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Introduction to SQL

1. What is a schema? How many schemas can be used in one database?

ANSWER: In the SQL environment, a schema is a logical group of database objects—such as tables and indexes—that are related to each other. Usually, the schema belongs to a single user or application. A singledatabase can hold multiple schemas that belong to different users or applications. Schemas are useful inthat they group tables by owner (or function) and enforce a first level of security by allowing each userto see only the tables that belong to that user.

PTS: 1 DIF: Difficulty: Moderate REF: p.251

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Definition Commands

1. Which command is used to save changes to the database? What is the syntax for this command?

ANSWER: Any changes made to the table contents are not saved on disk until a user closes the database, closesthe program he or she is using, or uses the COMMIT command. If the database is open and a poweroutage or some other interruption occurs before the user issues the COMMIT command, the user’schanges will be lost and only the original table contents will be retained. The syntax for the COMMITcommand is:

COMMIT [WORK]

The COMMIT command permanently saves all changes—such as rows added, attributes modified, androws deleted—made to any table in the database.

PTS: 1 DIF: Difficulty: Moderate REF: p.265-266

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. What is a subquery? When is it used? Does the RDBMS deal with subqueries any differently from normal queries?

ANSWER: A subquery, also known as a nested query or an inner query, is a query that is embedded (or nested)inside another query. The inner query is always executed first by the RDBMS.

In the SQL statement, INSERT INTO tablename SELECT columnlist FROM tableneme;, the INSERTportion represents the outer query, and the SELECT portion represents the subquery. A user can nestqueries (place queries inside queries) many levels deep; in every case, the output of the inner query isused as the input for the outer (higher-level) query.

PTS: 1 DIF: Difficulty: Moderate REF: p.270-271

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Data Manipulation Commands

1. What are the wildcard characters that are used with the LIKE command? Provide one or more examples of each.

ANSWER: The LIKE special operator is used in conjunction with wildcards to find patterns within string attributes.Standard SQL allows a user to use the percentage sign (%) and underscore (_) wildcard characters tomake matches when the entire string is not known:

% means any and all following or preceding characters are eligible. For example:

‘J%’ includes Johnson, Jones, Jernigan, July, and J-231Q.

‘Jo%’ includes Johnson and Jones.

‘%n’ includes Johnson and Jernigan.

_ means any one character may be substituted for the underscore. For example:‘_23-456-6789’ includes 123-456-6789, 223-456-6789, and 323-456-6789.

‘_23-_56-678_’ includes 123-156-6781, 123-256-6782, and 823-956-6788.

‘_o_es’ includes Jones, Cones, Cokes, totes, and roles.

PTS: 1 DIF: Difficulty: Moderate REF: p.280

NAT: BUSPROG: Technology STATE: DISC: Information Technology

KEY: Bloom’s: Knowledge TOP: Select Queries

1. How can a table be deleted from the database? Provide an example.

ANSWER: A table can be deleted from the database using the DROP TABLE command. For example, a user candelete the PART table with the following command:

DROP TABLE PART;

The user can drop a table only if it is not the “one” side of any relationship. If the user tries to drop atable otherwise, the RDBMS will generate an error message indicating that a foreign key integrityviolation has occurred.

PTS: 1 DIF: Difficulty: Moderate REF: p.290

NAT: BUSPROG: Analytic STATE: DISC: Information Technology

KEY: Bloom’s: Comprehension TOP: Additional Data Definition Commands