For the Multiple Choice Question you must select ALL correct options - not just

Question

For the Multiple Choice Question you must select ALL correct options - not just 1.

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 an id, 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 eld, nm, address,phone) . TICKET = {tld. eld, sld, timestamp, price, discount) This results in a database given by the following relations Ticket tld eld sld timestamp price discount 1 1 1 2018-03-23 3:14p 10S 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 10S none 513 2018-03-23 5:48pm 5$ lecturer Employe«e eld name 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 1 Nosferatu 2 Head On 2018-03-23 7pm Screen 1 2018-03-23 7pm Screen 2 3 Bloody Pit of Horror 2018-03-23 9pm Screen 1 2018-03-23 9pm Screen 2 The Following

Explanation / Answer

1)EMPLOYEE ,superkeys:

{eId,phone}, {name,phone}, {eId,name}

which are uniquely identifies the each record.

2) TICKET, superkeys:

{tID}

3)

(name)(EMPLOYEE) * sId(screen=="Screen 2")(Screening) *eID (Ticket)

4) name

Diane Selwyn

//returns the name from employee where the eID not in Ticket that, sId not in the Screening

any clarification, please do comments

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