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NEW QUESTION: 1
Sie führen eine SAP S / 4HANA-Migration durch. Einige der Buchungskreise wurden jahrelang NICHT verwendet und die zu diesen Buchungskreisen gebuchten Daten sind nicht mehr rechtlich gleichwertig.
Wie gehen Sie mit diesen Buchungskreisen um?
A. Migrieren Sie alle Buchungskreise, die Sie im Customizing ausgewählt haben.
B. Migrieren Sie alle Buchungskreise des Mandanten.
C. Alle Daten aus nicht verwendeten Buchungskreisen archivieren.
D. Alle als produktiv gekennzeichneten Buchungskreise migrieren.
Answer: B

NEW QUESTION: 2

A. Option B
B. Option C
C. Option D
D. Option A
Answer: B

NEW QUESTION: 3
The equation used to calculate the total number of symmetric keys (K) needed for a group of users
(N)
to communicate securely with each other is given by which of the following?
A. N(N - 1)/ 2
B. N(K - 1)/ 2
C. K(N + 1)/ 2
D. K(N - 1)/ 2
Answer: A
Explanation:
The formula is: Total number of users multiplied by total number of users minus 1,
the results are then divided by 2)
When using symmetric algorithms, the sender and receiver use the same key for encryption and
decryption functions. Each pair of users who want to exchange data using symmetric key
encryption must have two instances of the same key. This means that if Dan and Iqqi want to
communicate, both need to obtain a copy of the same key. If Dan also wants to communicate
using symmetric encryption with Norm and Dave, he needs to have three separate keys, one for
each friend. This might not sound like a big deal until Dan realizes that he may communicate with
hundreds of people over a period of several months, and keeping track and using the correct key
that corresponds to each specific receiver can become a daunting task.
If ten people needed to communicate securely with each other using symmetric keys, then 45 keys
would need to be kept track of. If 100 people were going to communicate, then 4,950 keys would
be involved.
The equation used to calculate the number of symmetric keys needed is N(N - 1)/ 2 = number of
keys
The following answers are incorrect:
K(N - 1)/ 2
N(K - 1)/ 2
K(N + 1)/ 2
The following reference(s) were/was used to create this question:
Harris, Shon (2012-10-18). CISSP All-in-One Exam Guide, 6th Edition (p. 782). McGraw-Hill.
Kindle Edition.

NEW QUESTION: 4
A customer has a web application with HTTP Sessions that need to be replicated to a backup site that is more than 100 miles away from the primary site, connected over the Internet.
Which type of session replication in WebLogic is recommended?
A. Database Replication
B. Synchronous Replication
C. Asynchronous Replication
D. WAN Replication
E. MAN Replication
Answer: C
Explanation:
Explanation/Reference:
WAN HTTP Session State Replication
Resources in a wide area network (WAN) are frequently spread across separate geographical regions. In addition to requiring network traffic to cross long distances, these resources are often separated by multiple routers and other network bottle necks. Network communication in a WAN generally has higher latency and slower interconnect.
Slower network performance within a WAN makes it difficult to use a synchronous replication mechanism like the one used within a MAN. WebLogic Server provides failover across clusters in WAN by using an asynchronous data replication scheme.
Note:
HTTP Session State Replication
Weblogic Server uses two methods for replicating HTTP session state across clusters:
in-memory replication
* Using in-memory replication, WebLogic Server copies a session state from one server instance to another. The primary server creates a primary session state on the server to which the client first connects, and a secondary replica on another WebLogic Server instance in the cluster. The replica is kept up-to-date so that it may be used if the server that hosts the servlet fails.
* JDBC-based persistence
In JDBC-based persistence, WebLogic Server maintains the HTTP session state of a servlet or JSP using file-based or JDBC-based persistence.
JDBC-based persistence is also used for HTTP session state replication within a Wide Area Network (WAN).