EBK MODERN DATABASE MANAGEMENT,
13th Edition
ISBN: 9780134792279
Author: TOPI
Publisher: YUZU
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Chapter 6, Problem 6.31PAE
Program Plan Intro
Find the relationships among the four given relations provided in figure. Name the primary keys and foreign keys belonging to each relation.
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Chapter 6 Solutions
EBK MODERN DATABASE MANAGEMENT,
Ch. 6 - Define each of the following terms: dynamic view...Ch. 6 - Match the following terms to the appropriate...Ch. 6 - Prob. 6.3RQCh. 6 - Prob. 6.4RQCh. 6 - Prob. 6.5RQCh. 6 - Explain the relationship between EXISTS and...Ch. 6 - Prob. 6.7RQCh. 6 - Under what conditions can a UNION clause be used?Ch. 6 - Prob. 6.9RQCh. 6 - Explain why it is necessary to limit the kinds of...
Ch. 6 - Describe a set of circumstances for which using a...Ch. 6 - Prob. 6.12RQCh. 6 - Prob. 6.13RQCh. 6 - Prob. 6.14RQCh. 6 - Prob. 6.15RQCh. 6 - Prob. 6.16RQCh. 6 - Prob. 6.17RQCh. 6 - Prob. 6.18RQCh. 6 - Prob. 6.19RQCh. 6 - List for advantages of SQL-invoked routines.Ch. 6 - Prob. 6.21RQCh. 6 - Prob. 6.22RQCh. 6 - Prob. 6.23RQCh. 6 - This chapter discusses the data dictionary views...Ch. 6 - Write an SQL query to answer the following...Ch. 6 - Write SQL retrieval commands for each of the...Ch. 6 - Write an SQL query to answer the following...Ch. 6 - Write SQL queries to answer the following...Ch. 6 - Write SQL queries to answer the following...Ch. 6 - Write an SQL query to answer the following...Ch. 6 - Prob. 6.31PAECh. 6 - Prob. 6.32PAECh. 6 - Prob. 6.33PAECh. 6 - What do you need to do if a tutor signs up and...Ch. 6 - Prob. 6.35PAECh. 6 - Write the SQL query to find any tutors who have...Ch. 6 - Prob. 6.37PAECh. 6 - Write an SQL query to determine the total number...Ch. 6 - Prob. 6.39PAECh. 6 - Prob. 6.40PAECh. 6 - Prob. 6.41PAECh. 6 - Prob. 6.42PAECh. 6 - Prob. 6.43PAECh. 6 - Which tutor needs to be reminded to tum in...Ch. 6 - Prob. 6.45PAECh. 6 - Write an SQL query to list all product line names...Ch. 6 - Modify to include only those product lines the...Ch. 6 - Prob. 6.48PAECh. 6 - Prob. 6.49PAECh. 6 - Write an SQL query to display the order number,...Ch. 6 - Prob. 6.51PAECh. 6 - Prob. 6.52PAECh. 6 - Prob. 6.53PAECh. 6 - Prob. 6.54PAECh. 6 - Prob. 6.55PAECh. 6 - Prob. 6.56PAECh. 6 - Prob. 6.57PAECh. 6 - Prob. 6.58PAECh. 6 - Write an SQL query to list each customer who...Ch. 6 - Prob. 6.60PAECh. 6 - Modify Problem and Exercise 6-60 so that the list...Ch. 6 - Prob. 6.62PAECh. 6 - Prob. 6.63PAECh. 6 - Prob. 6.64PAECh. 6 - Prob. 6.65PAECh. 6 - Prob. 6.66PAECh. 6 - Prob. 6.67PAECh. 6 - Prob. 6.68PAECh. 6 - Prob. 6.69PAECh. 6 - Prob. 6.70PAECh. 6 - Prob. 6.71PAECh. 6 - Rewrite your answer to Problem and Exercise 6-71...Ch. 6 - Display the customer ID, name, and order ID for...Ch. 6 - Prob. 6.74PAECh. 6 - Prob. 6.75PAECh. 6 - Prob. 6.76PAECh. 6 - Prob. 6.77PAECh. 6 - Prob. 6.78PAECh. 6 - Prob. 6.79PAECh. 6 - Write an SQL query to list the order number,...Ch. 6 - Prob. 6.81PAECh. 6 - Prob. 6.82PAECh. 6 - Prob. 6.83PAECh. 6 - Prob. 6.84PAECh. 6 - Prob. 6.85PAE
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- ERD for Sales Order Step 1. Draw an ER diagram for the story below. Use the ERD tool only. Draw Relations in Oracle notation. Please read the story carefully. Use the mentioned Entities ONLY. The company sells various Products to Customers. The Sales Agent books an Order. A new order is created for each Product. While booking an Order, Agent records Order Date, Product, Customer, Product Quantity, and Sales Agent. Product has Description and UnitPrice. Each Agent reports to a Manager. A Manger has a Nameand Commission Percent. NOTE: Total Amount is a calculation, doesn't belong to Order Step 2. Add Attributes to Sales Order ERD based on Rules of Normalization.arrow_forwardBased on the ER design create the schema for your online auction database, apply some combination of translating the ER components to relations and normalization. Provide SQL statements to create all of the relations in your database.arrow_forwardTo prevent modification anomalies, which of the following are created for a relation? Group of answer choices: 1. foreign keys 2. functional dependencies 3. referential integrity constraints 4. multi-valued attributes 5. primary keys 6. normal formsarrow_forward
- For each of the following relations, indicate the normal form for that relation. If the relation is not in third normal form, decompose it into 3NF relations. Functional dependencies are shown where appropriate. a) CLASS(CourseNo, SectionNo) b) CLASS(CourseNo, SectionNo, Room) c) CLASS(CourseNo, SectionNo, Room, Capacity) d) CLASS(CourseNo, SectionNo, CourseName, Room, Capacity)arrow_forwardBased on the following ER model and business rules, write an SQL statement to create COURSE and COURSE_PREREQ table. Be sure to implement all related data integrity constraint. For each column, use the data type as you think appropriate. Business rule • A course may have more than one prerequisites. The prerequisite courses may have their own prerequisite courses (more than one). COUSE entity • CourseNo is a UID (unique identifier). • Credit hours should be greater than 0 and less than 7. COURSE_PREREQ entity • UID is the composite of CourseNo and CourseNo (i.e., prerequisite courseNo). COURSE COURSE_PREREQ CourseNo: UID Title Description CourseNo: LCourseNo: UID CreditHoursarrow_forwardWrite the relations clearly showing the Primary key bolded and underlined (e.g., CustomerId). Foreign key shown in italics (e.g., OrderNumber) and separately defined beneath the relation (e.g., ShoppingCart OrderNumber FK to Order OrderNumber)arrow_forward
- 1. Follow the normalization process to normalize the above tables (you need not show the steps, but can use the blank sides of the exam if you want to work out the steps) and show the 3NF relational schema for each resulting table in the following format: ?NF TABLENAME (PRIMARY KEY, other attributes, FOREIGN KEY(s)) Use appropriate names for all tables and attributes. Do not draw any dependency diagrams. 2. Use the above relational schemas to draw the Crow’s Foot E-R Diagram (include all attributes, PKs, and FKs).arrow_forwardhow is the ProductID field in the Product table specified to be required? Why is it a required attribute?arrow_forwardThe value of the candidate key cannot uniquely identify each tuple in a relation. True False Question 1 A NULL value is a value that is not applicable or unknown. O True Falsearrow_forward
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