View the exhibit and examine the structure in ORDERS and ORDER_ITEMS tables.

You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATE VIEW statement would create the views successfully?

• A.CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) ||"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_dateWHITH CHECK OPTION;
• B.CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
• C.CREATE OR REPLACE VIEW ord_vu (order_id, order_date)AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
• D.CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;

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View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.

The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?

• A.UNION ALL
• B.UNION
• C.INTERSECT
• D.MINUS

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Examine these two queries and their output:
SELECT deptno, dname FROM dept;

SELECT ename, job, deptno FROM emp ORDER BY deptno;

Now examine this query:
SELECT ename, dname
FROM emp CROSS JOIN dept WHERE job = 'MANAGER'
AND dept.deptno IN (10, 20) ;

• A.12
• B.3
• C.64
• D.6

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Examine the business rule:
Each student can work on multiple projects and each project can have multiple students.
You need to design an Entity Relationship Model (ERD) for optimal data storage and allow for generating reports in this format:
STUDENT_ID FIRST_NAME LAST_NAME PROJECT_ID PROJECT_NAME PROJECT_TASK
Which two statements are true in this scenario? (Choose two.)

• A.The ERD must have a 1:M relationship between the STUDENTS and PROJECTS entities.
• B.The ERD must have a M:M relationship between the STUDENTS and PROJECTS entities that must be resolved into 1:M relationships.
• C.STUDENT_ID must be the primary key in the STUDENTS entity and foreign key in the PROJECTS entity.
• D.PROJECT_ID must be the primary key in the PROJECTS entity and foreign key in the STUDENTS entity.
• E.An associative table must be created with a composite key of STUDENT_ID and PROJECT_ID, which is the foreign key linked to the STUDENTS and PROJECTS entities.

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The CUSTOMERS table has a CUST_CREDT_LIMIT column of data type number.
Which two queries execute successfully?

• A.SELECT NVL2(cust_credit_limit * .15,'Not Available') FROM customers;
• B.SELECT TO_CHAR(NVL(cust_credit_limit * .15,'Not Available')) FROM customers;
• C.SELECT NVL2(cust_credit_limit,TO_CHAR(cust_credit_limit * .15),'NOT Available') FROM customers;
• D.SELECT NVL(cust_credit_limit * .15,'Not Available') FROM customers;
• E.SLECT NVL(TO_CHAR(cust_credit_limit * .15),'Not available') from customers;

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