Your query did not return the expected result; see below for further information.Your query resultYour query has returned the following result:namePeter SellersSigourney WeaverExpected resultThe expected result was the following:nameSigourney WeaverSusan SarandonDifference from the expected resultYour query did not return the following rows that are part of the expected result:nameSusan SarandonUnderstanding what went wrongYour query has been executed on the following database.Inspect the database to understand why your query did not return the expected result.Knowsid personA_id personB_id17 11 1219 11 1315 11 1418 12 1316 14 12Likesid personA_id personB_id19 11 1416 12 1118 13 1117 13 1215 14 12Personsid name address age eyeColor gender11 Woody allen Madrid 63 brown male12 Peter Sellers Amsterdam 19 brown male13 Susan Sarandon Paris 33 green female14 Sigourney Weaver Paris 24 blue female

You have a database with a customers table. Here is a partial listing of the records:First_NameLast_NameMichelleSenseAlexisRodriguezFloraSatoYou also have a table of employees. Here is a partial listing of the records:First_NameLast_NameSamuelClementineAlexisRodriguezJabariReddyYou want to create a query that will not return “Alexis Rodriguez” as part of the results. You begin typing your query:SELECT First_Name, Last_Name FROM customers________SELECT First_Name, Last_Name FROM employeesWhat should you type in the blank space between the two queries? (Please type your answer all in upper case. Example: SELECT)

This task concerns the following tables:PersonsidnameaddressageeyeColorgenderLikesidpersonA_id → PersonspersonB_id → PersonsKnowsidpersonA_id → PersonspersonB_id → PersonsThe semantics of the Likes table is that personA likes personB.The semantics of the Knows table is that personA knows personB.Write a query that returns the name of all persons that like everyone they know.

Find Customer RefereeGiven a table customer holding customers information and the referee.idnamereferee_id1WillNULL2JaneNULL3Alex24BillNULL5Zack16Mark2Write a query to return the list of customers NOT referred by the person with id'2'. For the sample data above, the result is:nameWillJaneBillZackOptionsselect namefrom customerwhere referee_id is null or referee_id <> 2;select namecustomerwhere referee_id is null or referee_id <> 2;select referee_idfrom customerwhere referee_id is null or referee_id <> 2;select namefrom customerwhere name is null or referee_id <> 2;

Consider the following database:MOVIE(id,title,yr)ACTOR(id,name)CASTING(movieid,actorid)Identify the SQL command which will return the titles of all 1959 Marilyn Monroe films.Option 1. The following SQL...SELECT title FROM movie,casting,actor WHERE movieid = movie.id<> AND name = 'Marilyn Monroe' ; Option 2. The following SQL...SELECT title FROM movie,actor WHERE name = 'Marilyn Monroe' AND yr = 1959 ; Option 3. The following SQL...SELECT title FROM movie,casting,actor WHERE movieid = movie.id AND actor.id = actorid AND name = 'Marilyn Monroe' AND yr = 1959 ; Option 4. The following SQL...SELECT title FROM movie,casting,actor WHERE movieid = movie.id AND actor.id = actorid AND movie.yr = casting.yr AND name = 'Marilyn Monroe' AND yr = 1959 ; Question 5Select one:a.Option 1b.Option 2c.Option 3d.Option 4

Question 4What will this query return?SELECT *FROM Books_tableLEFT JOIN Biography_table1 pointAll records in Books_table and any matching rows from Biography_tableAll records in both Books_table and Biography_tableAll records in Biography_table and any matching rows from Books_tableAll rows from Books_table joined together with Biography_table