Consider the following SQL query on the university schema:          select avg(salary)-(sum(salary) / count(*))           from instructor We might expect that the result of this query is zero since the average of a set of numbers is defined to be the sum of the numbers divided by the number of numbers. Indeed this is true for the example instructor

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Q: Take the university schema SQL query: select avg(salary) - (sum(salary)/count(*)) from teacher…

A: The AVG function calculates the AVG of the data by ignoring the null values in it. So, the possible…

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