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Note: \"Task 4\" is question 4, so its a 2 part question. Domain Consider a cine

ID: 3730109 • Letter: N

Question

Note: "Task 4" is question 4, so its a 2 part question.

Domain Consider a cinema database that stores ticket sales. Each Screening of a movie is stored in a relation with an id, the name of the movie, the time of the screening, and which screen it is shown on. An employee is stored with an id, the name, the address where the employee lives and the phone number. A ticket is stored with the id of the employee that sold the ticket, the id of the screening, the timestamp when it was sold, the price of the ticket and the discount type (if there was a discount). The database schema then is . SCREENING = {sld, movie, time, screen] . EMPLOYee- feld, name, address,phone) TICKET - feld, sId, timestamp, price, discount This results in a database given by the following relations Ticket tId eld sld timestamp price discount 1 1 1 2018-03-23 3:14pm 10 none 2 3 4 2018-03-23 3:16pm 8 student 3 3 2 2018-03-23 3:16pm 8 student 4 1 4 2018-03-23 5:48pm 10 none 5 1 3 2018-03-23 5:48pm lecturer Employee eld ame address phone 1 Tyler Durden 537 Paper Street(288) 555-0153 2 Diane Selwyn 1612 Havenhurst Dr. (255) 555-7614 3 Tyler Durden 1612 Havenhurst Dr. (255) 555-7614 Screening sld movie time screen 2018-03-23 7pm Screen 1 2018-03-23 7pm Screen 2 1 Nosferatu 2 Head On 3 Bloody Pit of Horror 2018-03-23 9pm Screen 1 4 The Following 2018-03-23 9pm Screen 2

Explanation / Answer

4. Result of the query :

The inner most query results in join of all tables where eID does not satisfy the conditions . So with NOT EXISTS, eID = 2 will be the result. So Name for eID = 2 will be the result of the query.

2. Relational Algebra

name ( Employee.eID != Ticket.eID and Ticket.sID != Screening.sID (Employee Ticket Screening))

The three tables are joined first and then conditions are applied resulting in Employee eID 2's name with projection operator.

Do ask if any query. Please upvote if the result is helpful.

   name Diane Selwyn
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