Databricks transcender data Analyst Databricks-certified-data-analyst-associate Exam Questions And Answers Pdf by Frazer q9 vce pdf

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Exam Name:	Databricks Certified Data Analyst Associate Exam
Exam Code:	Databricks-Certified-Data-Analyst-Associate Dumps
Vendor:	Databricks	Certification:	Data Analyst
Questions:	118 Q&A's	Shared By:	frazer

Question 20

The stakeholders.customers table has 15 columns and 3,000 rows of data. The following command is run:

Questions 20

After running SELECT * FROM stakeholders.eur_customers, 15 rows are returned. After the command executes completely, the user logs out of Databricks.

After logging back in two days later, what is the status of the stakeholders.eur_customers view?

Options:

The view remains available and SELECT * FROM stakeholders.eur_customers will execute correctly.

The view has been dropped.

The view is not available in the metastore, but the underlying data can be accessed with SELECT * FROM delta. `stakeholders.eur_customers`.

The view remains available but attempting to SELECT from it results in an empty result set because data in views are automatically deleted after logging out.

The view has been converted into a table.

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Question 21

A Data Analyst is working on employees_df and needs to add a new column where a 10% tax is calculated on the salary. Additionally, the DataFrame contains the column age, which is not needed.

Which code fragment adds the tax column and removes the age column?

Options:

employees_df = employees_df.withColumn( " tax " , employees_df.salary * 10).dropField( " age " )

employees_df = employees_df.withColumn( " tax " , employees_df.salary * 0.1).drop( " age " )

employees_df = employees_df.withColumn( " tax " , lit(0.1) * col( " salary " )).dropField( " age " )

employees_df = employees_df.withColumn( " tax " , employees_df.salary * 10).drop( " age " )

Discussion

Question 22

A data analyst wants the following output:

customer_name

number_of_orders

John Doe

388

Zhang San

234

Which statement will produce this output?

Options:

SELECT customer_name, count(order_id) AS number_of_ordersFROM customersJOIN ordersON customers.customer_id = orders.customer_idGROUP BY customer_name;

SELECT customer_name, count(order_id) number_of_ordersFROM customersJOIN ordersON customers.customer_id = orders.customer_id USE customer_name;

SELECT customerjiame, (order_id) number_of_ordersFROM customersJOIN ordersON customers.customer_id = orders.customer_id;

SELECT customerjiame, count(order_id)FROM customersJOIN ordersON customers.customer_id = orders.customer_id GROUP BY customerjiame;