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NEW QUESTION: 1
Wie leitet das System die Anforderungsart aus den Materialstammdaten ab?
A. MRP group -> strategy group -> planning strategy -> requirements type
B. Strategy group -> planning strategy -> requirements class -> requirements type
C. MRP group requirements class -> planning strategy -> requirements type
D. Strategy group -> planning strategy -> requirements type
Answer: A,D

NEW QUESTION: 2
Which two statements about the License Manager (LM) Configuration utility are true? (Choose two.)
A. The license file on the server can be edited where License Manager is installed.
B. The License Manager provides central control and administration of application licensing for all features of the Contact Center.
C. License Manager supports a WebLM License File and the Avaya WebLM Server on a virtual machine.
D. The License Manager Configuration utility can be used to download the license file from the Avaya license website.
Answer: B,C
Explanation:
Explanation/Reference:
Reference: https://downloads.avaya.com/css/P8/documents/101017335

NEW QUESTION: 3
View the Exhibit and examine the data 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 view successfully?

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