IDNameAge GenderEmplD CityDAhmed25Male3.Sara20Female43.Hassan31Male5.Raya27Female13CityEmploymentEmpIDCityIDCityNameDesignationProject ManagerSoftware EngineerMasqatSuhar223.Sur3Technician4Nizwa4AccountantSalalahSolve the following relational expressions for the above relations.a. IName( Age>25(Client))b. Old>2 or Age!=31(Client)C. OClient.EmpID3Employment.EmpID(Client X Employment)d. TIName, Gender(oCityName="Sur (User City))1234

Answered: Image /qna-images/answer/03227940-adb8-443e-a243-5a6f6a010c93.jpg

ethe following relational expressions for the above relations. TTName( Age>25(Client)) old>2 or Age!=31(Client) "Client.EmplD=Employment.EmplD(Client X Employment) TName, Gender(oCityName="Surr(User City))

ethe following relational expressions for the above relations. TTName( Age>25(Client)) old>2 or Age!=31(Client) "Client.EmplD=Employment.EmplD(Client X Employment) TName, Gender(oCityName="Surr(User City))

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Task 2: ' The Driver Relationship team wants to ensure that the all driving license IDs in the active drivers table have the length of 7. Alter the ACTIVE_DRIVERS table created in Chapter 8, Activity 1 to check the length of the DRIVER_DRIVING_LICENSE_ID.
Emp:- Id is Primary keyProject:- EmpId foreign key.Find out the output of the left outer join, right outer join and inner join.Take the below columns in output:-Emp table ID, name.Project table projectID, NAME
convert it to relations that are normalization
JOIN TABLES One row in the PAT_ENC table represents one patient encounter. One row in the ORDER MED table represents one medication order. One patient encounter can have many medication orders but one medication order can only have one patient encounter. In other words, the cardinality of this PAT_ENC to ORDER_MED relationship is one-to- many. You start a query with PAT_ENC. You then add ORDER_MED using an inner join. What is true about the granularity of the result set before and after adding the ORDER_MED table? SELECT ONE OF THE FOLLOWING A. The granularity stays at one row per patient encounter. B. The granularity stays at one row per medication order. C. The granularity changes from one row per patient encounter to one row per medication order on an encounter. D. The granularity changes from one row per medication order to one row per patient encounter.
PLZ HELP WITH THE FOLLOWING: make the following work select MOVIE_NAME, MOVIE_YEAR from MOVIE where MOVIE_COST > (select a.MOVIE_COST from MOVIE a, PRICE b where a.PRICE_CODE = b.PRICE_CODE and b.PRICE_DESC ="weekly Special"); CREATE TABLE MEMBERSHIP ( MEM_NUM CHAR(3) CONSTRAINT MEMBER_MEMNUM_PK PRIMARY KEY, MEM_FNAME VARCHAR(30) NOT NULL, MEM_LNAME VARCHAR(30) NOT NULL, MEM_STREET VARCHAR(30), MEM_CITY VARCHAR(10), MEM_STATE CHAR(2), MEM_ZIP CHAR(5), MEM_BALANCE NUMBER(2) ); CREATE TABLE RENTAL ( RENT_NUM CHAR(4) CONSTRAINT RENTAL_RENTNUM_PK PRIMARY KEY, RENT_DATE DATE, MEM_NUM CHAR(3), CONSTRAINT RENTAL_MEMNUM_FK FOREIGN KEY (MEM_NUM) REFERENCES MEMBERSHIP ); CREATE TABLE PRICE ( PRICE_CODE CHAR(1) CONSTRAINT PRICE_PRICECODE_PK PRIMARY KEY, PRICE_DESC VARCHAR(20), PRICE_RENTFEE NUMBER (3, 1), PRICE_DAILYATFEE NUMBER(3, 1) ); CREATE TABLE MOVIE ( MOVIE_NUM CHAR(4) CONSTRAINT MOVIE_MOVIENUM_PK PRIMARY KEY, MOVIE_NAME VARCHAR(30) NOT NULL,…
Please convert this EERD to relation. Thanks! Use arrow, not only box and tell where is the foreign key.
Please convert this EERD to relation. Thanks
Each relationship is mapped into a relation? true or false
Please convert this EERD to RELATION. Thanks! Use arrow to identify the foreign key, not only box and tell where is the foreign key. Draw it in paper, thanks!!!
Based on the ERD below, write the SQL statement to change the job description of “Database Designer” to “Database Developer”. After completing this task examine the results with a select statement. (Hint: Use an Update Statement.)
ProductID is the primary key column for the relation below. Which column violates a rule (characteristic) of a relation? Product Relation Name Cost ProductID Hammer 19.99 100 20 pack washers 6.75 200 Tape 3.19 300 1 gal paint 24.99 400 Group of answer choices The Cost column because it is out of order. The columns must be in alphabetical order. All the columns are valid. None of the columns are valid. The ProductID column. It is the primary key column and must be first.