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NEW QUESTION: 1
What is the primary goal of setting up a honeypot?
A. To entrap and track down possible hackers
B. To lure hackers into attacking unused systems
C. To set up a sacrificial lamb on the network
D. To know when certain types of attacks are in progress and to learn about attack techniques so the network can be fortified.
Answer: D
Explanation:
Explanation/Reference:
The primary purpose of a honeypot is to study the attack methods of an attacker for the purposes of understanding their methods and improving defenses.
"To lure hackers into attacking unused systems" is incorrect. Honeypots can serve as decoys but their primary purpose is to study the behaviors of attackers.
"To entrap and track down possible hackers" is incorrect. There are a host of legal issues around enticement vs entrapment but a good general rule is that entrapment is generally prohibited and evidence gathered in a scenario that could be considered as "entrapping" an attacker would not be admissible in a court of law.
"To set up a sacrificial lamb on the network" is incorrect. While a honeypot is a sort of sacrificial lamb and may attract attacks that might have been directed against production systems, its real purpose is to study the methods of attackers with the goals of better understanding and improving network defenses.
References
AIO3, p. 213

NEW QUESTION: 2
企業は現在のビジネスモデルを変更しています。変更には、複数のアジャイルチームを使用して複雑なプロジェクトを実装する必要があります。
プロジェクトマネージャーがテストチームにテストメカニズムとして採用させるべき3つのオプションは何ですか? (3つ選択してください)
A. ストーリーテスト
B. すべての可能な選択肢をカバーするための実験的アプローチ
C. 動作とテスト駆動開発に基づくテスト
D. セキュリティとパフォーマンスのテスト
E. 独立した個人テスト
Answer: A,C,D

NEW QUESTION: 3
You can record inspection results for an inspection lot, using defect data or characteristic results.
A. correct
B. Incorrect
Answer: A

NEW QUESTION: 4
Which two statements about the host address 172 150 100 10/18 are true?(Choose two)
A. The broadcast address is 172.150.127.255
B. The network address is 172.150.64.0
C. The network address is 172.150 0 0
D. The broadcast address is 172 150 255 255
F The broadcast address is 172.150.100 255
E. The network address is 172 150 100 0
Answer: A,B