Given the following table schemas (definitions) where "bid" is boat id, "brame" is boat name, "sname" is sailor name and "sid" is sailorid: Sailors(sid: integer, sname: string, rating: integer, age: real) Boats(bid: integer, bname: string, color: string) Reserves(sid: integer, bid: integer, day: date) Which one of the below given queries "Finds the names of sailors who have reserved two different boats on the same day. O a SELECT S.sname. FROM Sailors S, Reserves R1, Reserves R2 WHERE (S.sid-R1.sid OR S.sid-R2.sid) AND (R1.day <> R2.day AND R1.bid R2.bid) Ob SELECT S.sname. FROM Sailors S, Reserves R1, Reserves R2 WHERE R1.sid - R2.sid AND R1.day-R2.day AND R1.bid < R2.bid O SELECT S.sname. O FROM Sailors S, Reserves R1, Reserves R2 WHERE R1.sid-R2.sid AND R1.day-R2.day AND R1.bid=R2.bid Od. SELECT S.sname. FROM Sailors S, Reserves R1, Reserves R2 WHERE S.sid-R1.sid AND S.sid-R2.sid AND R1.day-R2.day AND R1.bid -R2.bid SELECT S.sname. FROM Sailors S, Reserves R1, Reserves R2 WHERE S.sid-R1.sid AND S.sid-R2.sid AND R1 day-R2.day AND R1.bid R2.bid

Q. Please write SQL codes to query data from member tables based on below criteria  Please write a SELECT statement to display each member’sdata that member_id is an odd number. Please display the result in ascending order of the last_name. The result should be identical to below results: [Screenshot]

How do you specify conditions for an Access query?

4 and 5

Create a table named company with these columns: Company_id varchar(255) Company_name varchar(255) default ‘x’ Hq_phone_number varchar(255) Primary key of company_id (please use constraint format) Unique key of Hq_phone_number (please use constraint format)

Q1- List all employees whose job_id is IT_PROG and hired after 01-JAN-95 and earning more than5000.Q2- List all employee except those with employee_id 101, 102, 103, 104 and 105Q3- Display the city attribute from locations table. Show the ones that contain “South” in the city.Sort the list in descending order.Q4- Display the first name, last name, salary, manager id and job id of all employees whose last fourcharacters of job id ends with “LERK”.Q5- Display the first name, last name and salary of all employees. Round the salary to the nearestunit of 1000 (ex. Salary=4500 New Salary=5000). Label the column as "New Salary".

Given the table PRODUCT(ProductID, ProductName, ProductClass, ProductWeight, ProductStyle, ProductColor, ProductPrice). Note: ProductWeight and ProductPrice are numeric fields, and the other attributes are character fields, i.e., store text. Write a SQL query to display the product class whose average price is at least 50

DATABASE

For this problem create a (temporary) table called instructor_course_nums. Write a procedure that accepts an instructor ID as input. The procedure calculates the total number of course sections taught by that instructor, and adds a tuple to the temporary table consisting of the instructors ID number, name, and total courses taught - call these attributes: ID, name, and tot_courses. If the instructor already has an entry in the table,then the procedure makes sure the total number of courses taught in the temporary table is up-to-date. You must name your procedure: Written in PostgreSQL

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,…