Printable 1Z0-061 Braindumps 2019

Exam Code: 1Z0-061 (1z0 061 dumps), Exam Name: Oracle Database 12c SQL Fundamentals, Certification Provider: Oracle Certifitcation, Free Today! Guaranteed Training- Pass 1Z0-061 Exam.

Free demo questions for Oracle 1Z0-061 Exam Dumps Below:

NEW QUESTION 1
You issued the following command: SQL> DROP TABLE employees; Which three statements are true?

  • A. All uncommitted transactions are committed.
  • B. All indexes and constraints defined on the table being dropped are also dropped.
  • C. Sequences used in the employees table become invalid.
  • D. The space used by the employees table is reclaimed immediately.
  • E. The employees table can be recovered using the rollback command.
  • F. The employees table is moved to the recycle bin.

Answer: ABF

Explanation: A: If a user issues a DDL (CREATE, ALTER, or DROP) or DCL (GRANT or REVOKE) command, the transaction in progress (if any) will be committed.
B: Dropping a table invalidates dependent objects, such as indexes and constraints. F: The DROP TABLE statement moves a table or object table to the recycle bin.
Incorrect:
Not B: In general sequences used in the table would not by affected when the table is dropped.
Not D: Unless you specify the PURGE clause, the DROP TABLE statement does not result in space being released back to the tablespace for use by other objects, and the space continues to count toward the user's space quota.
Not E: Dropping a table invalidates dependent objects and removes object privileges on the table. If you want to re-create the table, then you must regrant object privileges on the table, re-create the indexes, integrity constraints, and triggers for the table, and respecify its storage parameters.

NEW QUESTION 2
View the Exhibit and examine the structure of the PRODUCTS tables.
You want to generate a report that displays the average list price of product categories where the average list price is less than half the maximum in each category.
Which query would give the correct output?

  • A. SELECT prod_category, avg(prod_list_price)FROM productsGROUP BY prod_categoryHAVING avg(prod_list_price) < ALL(SELECT max(prod_list_price)/2FROM productsGROUP BY prod_category);
  • B. SELECT prod_category, avg(prod_list_price)FROM productsGROUP BY prod_categoryHAVING avg(prod_list_price) > ANY(SELECT max(prod_list_price)/2FROM productsGROUP BY prod_category);
  • C. SELECT prod_category, avg(prod_list_price)FROM productsHAVING avg(prod_list_price) < ALL(SELECT max(prod_list_price)/2FROM productsGROUP BY prod_category);
  • D. SELECT prod_category, avg(prod_list_price)FROM productsGROUP BY prod_categoryHAVING avg(prod_list_price) > ANY(SELECT max(prod_list_price)/2FROM products);

Answer: A

Explanation: Using the ANY Operator in Multiple-Row Subqueries
The ANY operator (and its synonym, the SOME operator) compares a value to each value returned by a subquery.
<ANY means less than the maximum.
>ANY means more than the minimum.
=ANY is equivalent to IN
Using the ALL Operator in Multiple-Row Subqueries
The ALL operator compares a value to every value returned by a subquery.
>ALL means more than the maximum and
<ALL means less than the minimum.
The NOT operator can be used with IN, ANY, and ALL operators.

NEW QUESTION 3
View the Exhibit and examine the structure of the CUSTOMERS table.
1Z0-061 dumps exhibit
In the CUSTOMERS table, the CUST_LAST_NAME column contains the values 'Anderson' and 'Ausson'.
You issue the following query:
1Z0-061 dumps exhibit
What would be the outcome?

  • A. 'Oder' and 'Aus'
  • B. an error because the TRIM function specified is not valid
  • C. an error because the LOWER function specified is not valid
  • D. an error because the REPLACE function specified is not valid

Answer: B

NEW QUESTION 4
The CUSTOMERS table has these columns:
The CUSTOMER_ID column is the primary key for the table. You need to determine how dispersed your customer base is.
Which expression finds the number of different countries represented in the CUSTOMERS table?

  • A. COUNT(UPPER(country_address))
  • B. COUNT(DIFF(UPPER(country_address)))
  • C. COUNT(UNIQUE(UPPER(country_address)))
  • D. COUNT DISTINCT UPPER(country_address)
  • E. COUNT(DISTINCT (UPPER(country_address)))

Answer: E

NEW QUESTION 5
Which best describes an inline view?

  • A. a schema object
  • B. a sub query that can contain an ORDER BY clause
  • C. another name for a view that contains group functions
  • D. a sub query that is part of the FROM clause of another query

Answer: D

Explanation: a sub query that is part of the FROM clause of another query
Incorrect
A- is not a schema object
B- sub query can contain GROUP BY clause as well.
C- does not necessary contains group functions
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 11-21

NEW QUESTION 6
View the Exhibit and examine the structure of CUSTOMERS table.
1Z0-061 dumps exhibit
Evaluate the following query:
1Z0-061 dumps exhibit
Which statement is true regarding the above query?

  • A. It executes successfully.
  • B. It produces an error because the condition on the CUST_CITY column is not valid.
  • C. It produces an error because the condition on the CUST_FIRST_NAME column is not valid.
  • D. It produces an error because conditions on the CUST_CREDIT_LIMIT column are not valid.

Answer: A

NEW QUESTION 7
Which three SQL statements would display the value 1890.55 as $1, 890.55? (Choose three.)

  • A. SELECT TO_CHAR(1890.55, '$0G000D00')FROM DUAL;
  • B. SELECT TO_CHAR(1890.55, '$9, 999V99')FROM DUAL;
  • C. SELECT TO_CHAR(1890.55, '$99, 999D99')FROM DUAL;
  • D. SELECT TO_CHAR(1890.55, '$99G999D00')FROM DUAL;
  • E. SELECT TO_CHAR(1890.55, '$99G999D99')FROM DUAL;

Answer: ADE

NEW QUESTION 8
Examine the structure proposed for the TRANSACTIONS table:
1Z0-061 dumps exhibit
Which two statements are true regarding the storage of data in the above table structure? (Choose two.)

  • A. The TRANS_DATE column would allow storage of dates only in the dd-mon-yyyy format.
  • B. The CUST_CREDIT_VALUE column would allow storage of positive and negative integers.
  • C. The TRANS_VALIDITY column would allow storage of a time interval in days, hours, minutes, and seconds.
  • D. The CUST_STATUS column would allow storage of data up to the maximum VARCHAR2 size of 4, 000 characters.

Answer: BD

Explanation: B: The NUMBER datatype stores fixed and floating-point numbers. Numbers of virtually any magnitude can be stored and are guaranteed portable among different systems operating Oracle, up to 38 digits of precision.
The following numbers can be stored in a NUMBER column:
Positive numbers in the range 1 x 10-130 to 9.99...9 x 10125 with up to 38 significant digits Negative numbers from -1 x 10-130 to 9.99...99 x 10125 with up to 38 significant digits Zero
Positive and negative infinity (generated only by importing from an Oracle Version 5 database)
D: The VARCHAR2 datatype stores variable-length character strings. When you create a table with a VARCHAR2 column, you specify a maximum string length (in bytes or characters) between 1 and 4000 bytes for the VARCHAR2 column.
An interval literal specifies a period of time, and Oracle supports two types of interval literals: YEAR_TO_MONTH and DAY TO SECOND. For DAY TO SECOND, you can specify these differences in terms in terms of days, hours, minutes, and seconds. DAY TO
SECOND contains a leading field and may contain an optional trailing field. If trailing field is specified it must be less significant than the leading field. For example, INTERVAL MINUTE TO DAY is not valid.
A DAY TO MINUTE interval considers an interval of days to the nearest minute. References:

NEW QUESTION 9
Examine the types and examples of relationships that follow:
1. One-to-one a) Teacher to students
2. One-to-many b) Employees to Manager
3. Many-to-one c) Person to SSN
4. Many-to-many d) Customers to products
Which option indicates the correctly matched relationships?

  • A. 1-a, 2-b, 3-c, and 4-d
  • B. 1-c, 2-d, 3-a, and 4-b
  • C. 1-c, 2-a, 3-b, and 4-d
  • D. 1-d, 2-b, 3-a, and 4-c

Answer: C

NEW QUESTION 10
See the Exhibit and examine the structure of ORD table: Exhibit:
1Z0-061 dumps exhibit
Evaluate the following SQL statements that are executed in a user session in the specified order:
CREATE SEQUENCE ord_seq; SELECT ord_seq.nextval FROM dual;
INSERT INTO ord
VALUES (ord_seq.CURRVAL, ’25-jan-2007, 101);
UPDATE ord
SET ord_no= ord_seq.NEXTVAL WHERE cust_id =101;
What would be the outcome of the above statements?

  • A. All the statements would execute successfully and the ORD_NO column would contain the value 2 for the CUST_ID 101.
  • B. The CREATE SEQUENCE command would not execute because the minimum value and maximum value for the sequence have not been specified.
  • C. The CREATE SEQUENCE command would not execute because the starting value of the sequence and the increment value have not been specified.
  • D. All the statements would execute successfully and the ORD_NO column would have the value 20 for the CUST_ID 101 because the default CACHE value is 20.

Answer: A

NEW QUESTION 11
Examine the structure of the EMPLOYEES table:
EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25)
Which three statements insert a row into the table? (Choose three.)

  • A. INSERT INTO employees VALUES ( NULL, 'John', 'Smith');
  • B. INSERT INTO employees( first_name, last_name) VALUES( 'John', 'Smith');
  • C. INSERT INTO employees VALUES ( 1000, 'John', NULL);
  • D. INSERT INTO employees (first_name, last_name, employee_id) VALUES ( 1000, 'John', 'Smith');
  • E. INSERT INTO employees (employee_id) VALUES (1000);
  • F. INSERT INTO employees (employee_id, first_name, last_name) VALUES ( 1000, 'John', ' ');

Answer: CEF

Explanation: EMPLOYEE_ID is a primary key.
Incorrect
A- EMPLOYEE_ID cannot be null
B- EMPLOYEE_ID cannot be null
D= mismatch of field_name with datatype
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 10-11

NEW QUESTION 12
Examine the structure of the EMP_DEPT_VU view:
1Z0-061 dumps exhibit
Which SQL statement produces an error?

  • A. SELECT *FROM emp_dept_vu;
  • B. SELECT department_id, SUM(salary) FROM emp_dept_vu GROUP BY department_id;
  • C. SELECT department_id, job_id, AVG(salary) FROM emp_dept_vu GROUP BY department_id, job_id;
  • D. SELECT job_id, SUM(salary) FROM emp_dept_vu WHERE department_id IN (10, 20) GROUP BY job_id HAVING SUM(salary) > 20000;
  • E. None of the statements produce an error; all are valid.

Answer: E

Explanation: None of the statements produce an error.
Incorrect
A- Statement will not cause error
B- Statement will not cause error
C- Statement will not cause error
D- Statement will not cause error

NEW QUESTION 13
Which two statements are true regarding views? (Choose two.)

  • A. A simple view in which column aliases have been used cannot be updated.
  • B. Rows cannot be deleted through a view if the view definition contains the DISTINCT keyword.
  • C. Rows added through a view are deleted from the table automatically when the view is dropped.
  • D. The OR REPLACE option is used to change the definition of an existing view without dropping and recreating it.
  • E. The WITH CHECK OPTION constraint can be used in a view definition to restrict the columns displayed through the view.

Answer: BD

NEW QUESTION 14
View the Exhibit and examine the structure of the PROMOTIONS, SALES, and CUSTOMER tables.
You need to generate a report showing the promo name along with the customer name for all products that were sold during their promo campaign and before 30th October 2007.
You issue the following query:
1Z0-061 dumps exhibit
Which statement is true regarding the above query?

  • A. It executes successfully and gives the required result.
  • B. It executes successfully but does not give the required result.
  • C. It produces an error because the join order of the tables is incorrect.
  • D. It produces an error because equijoin and nonequijoin conditions cannot be used in the same SELECT statement.

Answer: B

NEW QUESTION 15
Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables:
1Z0-061 dumps exhibit
Which MERGE statement is valid?

  • A. MERGE INTO new_employees c USING employees e ON (c.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET name = e.first_name ||', '|| e.last_name WHEN NOT MATCHED THEN INSERT value S(e.employee_id, e.first_name||', '||e.last_name);
  • B. MERGE new_employees c USING employees e ON (c.employee_id = e.employee_id) WHEN EXISTS THEN UPDATE SET name = e.first_name ||', '|| e.last_name WHEN NOT MATCHED THEN INSERT valueS(e.employee_id, e.first_name ||', '||e.last_name);
  • C. MERGE INTO new_employees cUSING employees e ON (c.employee_id = e.employee_id) WHEN EXISTS THEN UPDATE SET
  • D. name = e.first_name ||', '|| e.last_name WHEN NOT MATCHED THEN INSERT value S(e.employee_id, e.first_name ||', '||e.last_name);
  • E. MERGE new_employees c FROM employees e ON (c.employee_id = e.employee_id) WHEN MATCHED THEN UPDATE SET name = e.first_name ||', '|| e.last_name WHEN NOT MATCHED THEN INSERT INTO new_employees valueS(e.employee_id, e.first_name ||', '||e.last_name);

Answer: A

Explanation: this is the correct MERGE statement syntax
Incorrect
B- it should MERGE INTO table_name
C- it should be WHEN MATCHED THEN
D- it should MERGE INTO table_name
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 8-29

NEW QUESTION 16
Which normal form is a table in if it has no multi-valued attributes and no partial dependencies?

  • A. First normal form
  • B. Second normal form
  • C. Third normal form
  • D. Fourth normal form

Answer: B

Explanation: According to the Second Normal Form (2NF) there must be no partial dependencies on a concatenated key.

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