19. Draw the query tree for the SQL statement below, then draw the optimized heuristic tree. PID is the Primary key of Provider e SELECT Provider.Name, Networking.Name, Show.Name, Show.Description FROM Provider, Network, Show WHERE Provider.PID = Network.PID AND Show.NetworklD = Network.NetworkID AND Show.Rating > 4 AND Provider.PID = 456

Implement a new independent entity phone in the Sakila database. Attributes and relationships are shown in the following diagram: The diagram uses Sakila naming conventions. Follow the Sakila conventions for your table and column names: All lower case Underscore separator between root and suffix Foreign keys have the same name as referenced primary key Write CREATE TABLE and ALTER TABLE statements that: Implement the entity as a new phone table. Implement the has relationships as foreign keys in the Sakila customer, staff, and store tables. Remove the existing phone column from the Sakila address table. Step 2 requires adding a foreign key constraint to an existing table. Ex: ALTER TABLE customer ADD FOREIGN KEY (phone_id) REFERENCES phone(phone_id) ON DELETE SET NULL ON UPDATE CASCADE; Specify data types as follows: phone_id, phone_number, and country_code have data type INT. phone_type has date type VARCHAR(12) and contains strings like 'Home', 'Mobile', and 'Other'. Apply…

Consider the following schema:instructor(ID, name, dept_name, salary)department(dept_name, building, college, budget)What is the correct SQL statement to perform a query to find the names and average salaries of departments in the college of ‘Science’ whose average salary is greater than 75000? (a) select dept_name, avg (salary)from instructor natual join departmenthaving college = ‘Science’group by dept_namewhere avg (salary) > 75000;(b) select dept_name, avg (salary)from instructor natual join departmentwhere college = ‘Science’group by dept_namewhere avg (salary) > 75000;(c) select dept_name, avg (salary)from instructor natual join departmentwhere college = ‘Science’group by dept_namehaving avg (salary) > 75000;(d) select dept_name, avg (salary)from instructor natual join departmenthaving college = ‘Science’group by dept_namehaving avg (salary) > 75000;(e) select dept_name, avg (salary)from instructor natual join departmentwhere college = ‘Science’and avg (salary) >…

Write an SQL statement to count the number of distinct breeds.The following table schema for the PET_3 table is another alternate version of the PETtable:PET_3 (PetID, PetName, PetType, PetBreed, PetDOB, PetWeight, OwnerID)Data for PET_3 are shown in Figure 3-20. Except as specifically noted in the question itself, usethe PET_3 table for your answers to all the remaining review questions.