Job-Ready SQL in an Afternoon - Aggregation Functions

Job-Ready SQL in an Afternoon - Aggregation Functions

Assessment

Interactive Video

Information Technology (IT), Architecture, Physics, Science

University

Hard

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The video tutorial explores various SQL aggregation functions, including AVG, MAX, MIN, SUM, and COUNT, to analyze data in a track table. It demonstrates how to calculate average, maximum, and minimum durations, sum total times, and count instances, while addressing null values. Additionally, it covers converting milliseconds to fractional minutes using arithmetic operations.

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7 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the primary purpose of using aggregations in SQL?

To modify the structure of the database

To create new tables

To perform calculations on data sets

To delete data from tables

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which SQL function is used to find the average value of a column?

AVG

MIN

COUNT

SUM

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How can you find the longest time in a column using SQL?

Using the MIN function

Using the COUNT function

Using the MAX function

Using the AVG function

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the COUNT function do in SQL?

It counts the number of occurrences in a column

It finds the average of a column

It sums up all values in a column

It finds the maximum value in a column

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do null values affect the COUNT function in SQL?

They are included in the count

They are ignored in the count

They cause an error

They double the count

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is one way to count the total number of rows in a table?

Using the MIN function

Using the MAX function

Using a star (*) in the COUNT function

Using the AVG function

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How can you convert milliseconds to minutes in SQL?

By multiplying by 1000 and then by 60

By adding 1000 and then 60

By subtracting 1000 and then 60

By dividing by 1000 and then by 60