1Z0-071 NEW QUESTION & EXAM 1Z0-071 SAMPLE

1z0-071 New Question & Exam 1z0-071 Sample

1z0-071 New Question & Exam 1z0-071 Sample

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Oracle 1z1-071 exam is a computer-based exam that consists of 73 multiple-choice questions. Candidates have two hours to complete the exam and must achieve a passing score of 63% or higher to earn the certification. 1z0-071 exam can be taken at any Oracle testing center or through Pearson VUE testing centers worldwide.

Oracle 1z1-071 certification exam covers a broad range of topics that are considered essential for anyone seeking to become an expert in database management. 1z0-071 Exam Tests an individual's skill in designing, manipulating, retrieving, and managing SQL database applications. It also emphasizes the ability to utilize advanced SQL features such as subqueries, data validation, and data manipulation.

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Exam 1z0-071 Sample, Sample 1z0-071 Questions

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Oracle 1z0-071 certification exam covers various topics, including SQL fundamentals, data retrieval using SQL select statements, data manipulation using SQL DML statements, database object creation and management, and control statements. 1z0-071 exam is designed to challenge the candidate's understanding of SQL and its use in managing and manipulating data in an Oracle database environment. 1z0-071 Exam requires the candidate to have a thorough understanding of SQL concepts and their practical application in real-world scenarios.

Oracle Database SQL Sample Questions (Q238-Q243):

NEW QUESTION # 238
Refer to the exhibit.

Which two queries only return CUBE?

  • A. SELECT shape FROM bricks JOIN boxes ON weight < max_weight;
  • B. SELECT shape FROM bricks JOIN boxes ON NOT (weight > max_weight);
  • C. SELECT shape FROM bricks JOIN boxes ON weight >= min_weight AND weight < max_weight;
  • D. SELECT shape FROM bricks JOIN boxes ON weight > min_weight;
  • E. SELECT shape FROM bricks JOIN boxes ON weight BETWEEN min_weight AND max_weight;

Answer: B,C

Explanation:
Based on the table structure given in the image, to return the value 'CUBE' from the 'bricks' table when joined with 'boxes', the condition must ensure that the weight of the bricks is within the allowed weight range specified in the 'boxes' table for a 'SMALL' box size.
A . True. Since MAX_WEIGHT is 0, a comparison using >= min_weight AND weight < max_weight will only return rows where the weight is less than 0, which is impossible for actual weight values, suggesting there might be a mistake in the data provided or the comparison logic.
E . True. NOT (weight > max_weight) effectively translates to 'where weight is less than or equal to max_weight'. However, since MAX_WEIGHT is 0, this condition would only be true if the weight is not greater than 0, which can only happen if the weight is 0 or less. This seems to indicate an anomaly where either the data is incorrect, or the condition is meant to handle a case where the weight is zero or possibly a negative placeholder value.
Both B and D will potentially return more than just 'CUBE' if there are bricks with weights greater than MIN_WEIGHT. C is incorrect because BETWEEN is inclusive, and there are no weights that are both greater than or equal to MIN_WEIGHT and less than or equal to MAX_WEIGHT when MAX_WEIGHT is 0.


NEW QUESTION # 239
View the exhibit and examine the structure of the EMPLOYEES table.

You want to display all employees and their managers having 100 as the MANAGER_ID.
You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

  • A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id = 100;
  • B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id = 100;
  • C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100
  • D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id = 100;

Answer: D


NEW QUESTION # 240
Examine the description of the PROMTIONS table:

You want to display the unique promotion costs in each promotion category.
Which two queries can be used?

  • A. SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;
  • B. SELECT promo_category, DISTINCT promo_cost PROM promotions ORDER BY 2:
  • C. SELECT DISTINCT promo_cost ||' in' II DISTINCT promo_category FROM promotions ORDER BY 1;
  • D. SELECT promo_cost, | pxomo_category FROM promotions ORDER BY 1;
  • E. SELECT DISTINCT promo_category ||'has'|| promo_cost AS COSTS FROM promotions ORDER BY 1;

Answer: A,E

Explanation:
To display unique promotion costs in each promotion category, the correct queries that can be used are:
C . SELECT DISTINCT promo_category ||' has '|| promo_cost AS COSTS FROM promotions ORDER BY 1; This query concatenates the promo_category with the literal ' has ' and promo_cost, giving a unique string for each combination of promo_category and promo_cost, which is what we are interested in when we want to list unique costs per category.
D . SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1; This query selects distinct combinations of promo_category and promo_cost, which is exactly what's required to display unique promotion costs in each category.
Options A, B, and E are incorrect:
A is incorrect because it does not use the DISTINCT keyword to ensure uniqueness.
B has incorrect syntax; the DISTINCT keyword should appear directly after SELECT and applies to all columns, not just one.
E is syntactically incorrect and would not execute because the DISTINCT keyword is not used correctly. It should not appear twice in the SELECT clause.


NEW QUESTION # 241
Examine the structure of the two tables.

Which two queries execute successfully? (Choose two.)

  • A. Option E
  • B. Option A
  • C. Option B
  • D. Option C
  • E. Option D

Answer: D,E


NEW QUESTION # 242
View the Exhibit and examine the structure of the PRODUCT_INFORMATION table.

You want to see the product names and the date of expiration of warranty for all the products, if the product is purchased today. The products that have no warranty should be displayed at the top and the products with maximum warranty period should be displayed at the bottom.
Which SQL statement would you execute to fulfill this requirement?

  • A. SELECT product_name, SYSDATE+warranty_period AS "Warranty expire date"FROM product_informationORDER BY SYSDATE+warranty_period
  • B. SELECT product_name, SYSDATE+warranty_period AS "Warranty expire date"FROM product_informationORDER BY SYSDATE-warranty_period
  • C. SELECT product_name, SYSDATE+warranty_period AS "Warranty expire date"FROM product_informationORDER BY SYSDATE
  • D. SELECT product_name, SYSDATE+warranty_period AS "Warranty expire date"FROM product_informationWHERE warranty_period > SYSDATE

Answer: A


NEW QUESTION # 243
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P.S. Free 2025 Oracle 1z0-071 dumps are available on Google Drive shared by ExamTorrent: https://drive.google.com/open?id=10M5yuUwAChUcup_a8tuWOHWKzYxQ-cIP

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