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Vendor: The Open Group

Exam Code: OGA-032 Dumps

Questions and Answers: 104

Product Price: $69.00

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NEW QUESTION: 1
Welche der folgenden Online-Auditing-Techniken ist am effektivsten, um Fehler oder Unregelmäßigkeiten frühzeitig zu erkennen?
A. Eingebettetes Überwachungsmodul
B. Audit Hooks
C. Schnappschüsse
D. Integrierte Testeinrichtung
Answer: B
Explanation:
Erläuterung:
Bei der Audit-Hook-Technik wird Code zur Prüfung ausgewählter Transaktionen in Anwendungssysteme eingebettet. Dies hilft einem IS-Prüfer zu handeln, bevor ein Fehler oder eine Unregelmäßigkeit außer Kontrolle gerät. Bei einem eingebetteten Prüfmodul wird speziell geschriebene Software in das Host-Anwendungssystem des Unternehmens eingebettet, sodass Anwendungssysteme selektiv überwacht werden. Eine integrierte Testeinrichtung wird verwendet, wenn die Verwendung von Testdaten nicht praktikabel ist, und Schnappschüsse werden verwendet, wenn ein Audittrail erforderlich ist.

NEW QUESTION: 2
DRAG DROP
You develop a live streaming solution using Azure Media Services.
You need to configure live Streaming tor single bitrate encoding. All recorded content must be retained for 20 hours.
Which three actions should you perform in sequence? To answer, move all actions from the list of actions to the answer area and arrange them in the correct order.

Answer:
Explanation:


NEW QUESTION: 3
ユーザーは、許可されたシステムのMACアドレスをスプーフィングすることにより、誰かが許可なくホームネットワークにアクセスしている疑いがあります。現在ホームネットワーク上にあります。

ネットワーク上に不正なユーザーがいるかどうかを判断するための次のステップは、次のうちどれですか?
A. MACフィルタリングを適用し、ルーターがシステムのいずれかをドロップするかどうかを確認します。
B. 承認された各システムを物理的にチェックして、ネットワークに切り替えられているかどうかを判断します。
C. 許可された各システムのpingスイープを実行し、エコー応答が受信されるかどうかを確認します。
D. ホスト名が不明であり、MACフィルタリングがこのホストに適用されないため、「不明な」ホストを拒否します。
Answer: B
Explanation:
AまたはCにすることはできません。質問では、MACアドレスがスプーフィングされており、フィルターをバイパスできることが示唆されているためです。
Dは無意味です。

NEW QUESTION: 4
Assuming the port statements are correct, which two (three?) code fragments create a one-byte file?
A. OutputStream fos = new FileOutputStream(new File("/tmp/data.bin"));
OutputStream bos = new BufferedOutputStream(fos);
DataOutputStream dos = new DataOutputStream(bos);
dos.writeByte(0);
dos.close();
B. OutputStream fos = new FileOutputStream ("/tmp/data.bin"); fos.writeByte(0); fos.close();
C. OutputStream fos = new FileOutputStream (new File ("/tmp/data.bin"));
DataOutputStream dos = new DataOutputStream(fos);
dos.writeByte(0);
dos.close();
D. OutputStream fos = new FileOutputStream ("/tmp/data.bin"); DataOutputStream dos = new DataOutputStream(fos); dos.writeByte(0); dos.close();
Answer: A,C,D
Explanation:
B:Create DataOutputStream from FileOutputStream public static void main(String[] args) throws
Exception { FileOutputStream fos = new FileOutputS tream("C:/demo.txt"); DataOutputStream dos
= new
DataOutputStream(fos);
Note:
The FileOutputStream class is a subclass of OutputStream. You can construct a FileOutputStream
object by
passing a string containing a path name or a File object.
You can also specify whether you want to append the output to an existing file.
public FileOutputStream (String path)
public FileOutputStream (String path, boolean append)
public FileOutputStream (File file)
public FileOutputStream (File file, boolean append)
With the first and third constructors, if a file by the specified name already exists, the file will be
overwritten. To append to an existing file, pass true to the second or fourth constructor.
Note 2:public class DataOutputStream extends FilterOutputStream implements DataOutput
A data output stream lets an application write primitive Java data types to an output stream in a
portable way.
An application can then use a data input stream to read the data back in.
Reference:java.io Class DataOutputStream


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