The following code would include: SELECT P.P_DESCRIPT, P.V_CODE, V.V_NAME FROM PRODUCT P RIGHT OUTER JOIN VENDOR V ON P.V_CODE=V.V_CODE Group of answer choices   A) all rows of the PRODUCT Table regardless of matches with the VENDOR Table. B) all rows of the VENDOR Table regardless of matches with the PRODUCT Table. C) only rows that match both PRODUCT and VENDOR Tables

1.What is the result when the following SQL query is executed on a table named products? Choose the most correct answer. SELECT AVG(Products)FROM Products; Group of answer choices The query results in an error because the is no AVG function None of the above The query results in an error be cause a column is not specified The query returns the average of the Products table   2. If the primary key in the Employee table is located in the "ID" column what is the maximum number of results the following SQL SELECT query will return? SELECT *FROM EmployeeWHERE ID = 15; Group of answer choices 2 1 3 Do not have enough information to determine the answer 0

USE this part to write SQL statements at the bottom questions 1-11   Branch(branch_id:integer, branch_name:varchar(50), branch_location:varchar(40), money_on_hand:numeric(15,2)       create table Branch             (branch_id                              integer,     branch_name                                 varchar(50),     branch_location                 varchar(40),     money_on_hand                 numeric(15,2),     primary key (branch_id));   Loan(loan_number:integer, branch_id:integer, amount:numeric(8,2))                   foreign key branch_id references Branch(branch_id)   create table Loan       (loan_number            integer,     branch_id                                  integer,     amount                                     numeric(8,2),     primary key (loan_number),     foreign key (branch_id) references Branch (branch_id));   Customer(customer_id:integer, customer_last_name:varchar(35),customer_first_name:varchar(25), customer_street:varchar(30),…

Display the country names and region names where no warehouse is located in them. Use set operators to create this report and provide the exact sorting as below.

write following queries

Given the following schema, answer the following questions    USERS Attribute Data Type USER_ID NUMBER(3) USER_NAME VARCHAR2(20) USER_TYPE VARCHAR2(10) USER_LIMIT NUMBER(1)                           BOOKS Attribute Data Type BOOK_ID NUMBER(5) BOOK_TITLE VARCHAR2(25) BOOK_YEAR VARCHAR2(4)       Write the SQL code that will create the tables structures for the tables named Books, and Users. Identify and include any PK and FK for both tables.

Topic 1: The 5 JoinsRun each of the following statements in APEX. Explain the differences in results.SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES EINNER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ELEFT OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ERIGHT OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES EFULL OUTER JOIN HOL_JOBS J ON E.Job_ID=J.Job_ID;SELECTE.Last_Name,J.Job_TitleFROM HOL_EMPLOYEES ECROSS JOIN HOL_JOBS J;                 DBMS 130 Worksheet: JOINS and SET Operations2Topic 2: Write an SQL cartesian product between Regions and Countries.Copy your APEX here.How many rows are in the result set?How could you have calculated the number of rows in the result set BEFORE running yourquery?Topic 3: Briefly explain how set operators handle NULL values.Run the following 3 SQL queries in APEX which use set operators over these…

Question 3: Create the DDL statements to create the tables described in your modified ER diagram in question 2: DDL Statement: CREATE TABLE School ( school_code integer Primary Key(10), location varchar(25), name varchar(10) ); CREATE TABLE Professors ( staff_id integer(10), name varchar(20), school varchar(30) ); CREATE TABLE Course ( subject_code varchar(10), school varchar(30), name varchar(10) ); CREATE TABLE Subject ( subject_code(10), school varchar(30), name varchar(10) ); CREATE TABLE Student ( student_id number(10), name varchar(10), address varchar(30), phone number(10) ); Table Description desc table School; Select * from School; Name                       Null              Type school_code Not null integer location Not null varchar(25) name Not null varchar(10)   desc table Professor; Select * from Professor; Name               Null              Type staff_id Not null number(10) name   varchar(20) school   varchar(30)   desc table Course; Select * from…