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Exam Code: Data-Engineer-Associate Dumps

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NEW QUESTION: 1
Click the Exhibit button.

In the exhibit, R2 and R3 are both rendezvous points. Assume that R2 fails. Which RP redundancy method could converge the multicast stream and RP as quickly as the IGP?
A. Auto-RP in combination with MSDP
B. Anycast RP and MSDP
C. Auto-RP without using MSDP
D. BSR without the use of MSDP
Answer: B

NEW QUESTION: 2



A. ALTER TABLE Orders
ADD CONSTRAINT CK_Orders_CustomerID
CHECK (CustomerId IN (SELECT CustomerId FROM Customer))
B. ALTER TABLE Orders
ADD CONSTRAINT FX_Orders_CustomerID FOREIGN KEY (CustomerId) REFERENCES
Customer (CustomerId)
C. ALTER TABLE Orders
ADD CONSTRAINT PK Orders CustomerId PRIMARY KEY (CustomerID)
D. ALTER TABLE Customer
ADD OrderId INT NOT NULL;
ALTER TABLE Customer
ADD CONSTRAINT FK_Customer_OrderID FOREIGN KEY (OrderID) REFERENCES
Orders
(OrderID);
E. ALTER TABLE Customer
ADD CONSTRAINT FK_Customer_CustomerID FOREIGN KEY {CustomerID)
REFERENCES
Orders (CustomerId)
Answer: B
Explanation:
Reference: http://msdn.microsoft.com/en-us/library/ms189049.aspx

NEW QUESTION: 3
You have an Exchange Server 2016 organization. The organization contains 5,000 mailboxes.
All users connect to their mailbox by using Microsoft Outlook and Outlook on the web.
You need to ensure that all of the users are notified before they send an email message to a distribution group named Executives.
Which cmdlet should you use?
A. Set-Mailbox
B. Set-DLPPolicy
C. Set-DistributionGroup
D. New-TransportRule
E. New-PolicyTipConfig
Answer: D
Explanation:
Explanation/Reference:
Explanation:
You can use a transport rule with the condition The recipient is a member of. This condition matches messages that contain recipients who are members of the specified distribution group. The distribution group can be listed in the To, Cc, or Bcc fields.

NEW QUESTION: 4
View the exhibit and examine the structure of ORDERS and CUSTOMERS tables.

Which INSERT statement should be used to add a row into the ORDERS table for the customer whose CUST_LAST_NAME is Roberts and CREDIT_LIMIT is 600? Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.
A. INSERT INTO ordersVALUES (1,'10-mar-2007', 'direct',(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
B. INSERT INTO orders (order_id, order_date, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), order_total)VALUES (1,'10-mar-2007', 'direct', &&customer_id, 1000);
C. INSERT INTO orders (order_id, order_date, order_mode,(SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), order_total)VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);
D. INSERT INTO(SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_totalFROM orders o, customers cWHERE o.customer_id = c.customer_id AND c.cust_last_name='Roberts' AND c.credit_limit=600)VALUES (1,'10-mar-2007', 'direct', (SELECT customer_idFROM customersWHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
Answer: A


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