We want to find the median of a 4-element array, that is, the 2nd smallest element in the array. V the minimum number of comparisons that you need to get the solution?

Write and execute the command to retrieve the customer ID, first name, and last name of each customer in the MAJOR_CUSTOMER VIEW with a balance that exceeds the credit limit.

When is creating an index manually inappropriate?a. when queries return a large percentage of rows in the resultsb. when the table is smallc. when the majority of table operations are updatesd. all of the abovee. only a and c

In MySQL please: The Question: Create a trigger named trg_char_hours that automatically updates the AIRCRAFT table when a new CHARTER row is added. Use the CHARTER table’s CHAR_HOURS_FLOWN to update the AIRCRAFT table’s AC_TTAF, AC_TTEL, and AC_TTER values. (Hint: use temp values, as the INSERT event does not accept the OLD keyword.)

Without using the MAJOR_CUSTOMER VIEW, retrieve the customer ID, first name, and last name for every customer whose credit limit is $500 or less.

Which result set requires a JOIN?     Showing all album titles with both artist ID and album ID Showing artist names with artist_id's Showing artist_id with album title Showing all album_id's with artist names

Produce a script file with 15 queries: At least two using join At least three using where clause At least three using order by At least two using single-row functions At least two using group by and group functions at least one nested query At least 4 queries from chapter 1 - 6 (exclude Create Table, Insert queries) ------- TABLE 1 create TABLE ProCoaches(pro_co_ID INTEGER PRIMARY KEY,pro_coachName Varchar(50) NOT NULL,pro_coachLocation Varchar(20)); INSERT INTO ProCoaches VALUES (1, 'John', 'Missoula'); INSERT INTO ProCoaches VALUES (2, 'Joe', 'Bozeman');INSERT INTO ProCoaches VALUES (3, 'Tara','Butte'); select * FROM ProCoaches;select LENGTH('Missoula') AS LengthOfLocation;select pro_coachName FROM ProCoaches where pro_coachLocation='Missoula'; select COUNT(pro_co_ID), pro_coachLocation FROM ProCoaches GROUP BY pro_coachLocation; TABLE 2 create TABLE ProCyclist(pro_cy_ID VARCHAR(10) PRIMARY KEY,pro_cyclist_Name VARCHAR(50) NOT NULL,pro_cyclist_Gender Varchar(20),pro_cyclist_DOB…