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Certified Fraud Examiner - Financial Transactions and Fraud Schemes Exam
- Exam Number/Code : CFE-Financial-Transactions-and-Fraud-Schemes
- Exam Name : Certified Fraud Examiner - Financial Transactions and Fraud Schemes Exam
- Questions and Answers : 213 Q&As
- Update Time: 2019-01-10
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NEW QUESTION: 1
Ferris Plastics、Inc.は中規模企業で、ユーザーPCから企業サーバーへのLAN接続を提供するエンタープライズネットワーク(アクセス、ディストリビューション、コアスイッチ)を備えています。 分散スイッチは、HSRPを使用して高可用性ソリューションを提供するように設定されています。
* DSW1 - VLAN 101 VLAN 102およびVLAN 105用のプライマリデバイス
* DSW2 - VLAN 103およびVLAN 104のプライマリデバイス
* プライマリデバイスのGigabitEthemet1 / 0/1に障害が発生すると、バックアップデバイスのGigabitEthernet1 / 0/1も障害が発生していない限り、プライマリデバイスのステータスがプライマリデバイスとして解放されます。
トラブルシューティングではいくつかの問題が特定されています。 現在、すべてのインターフェースが稼動しています。 実行コンフィギュレーションとshowコマンドを使用して、次の質問を調査して対応するよう求められました。
DSW2上のVLAN 105 HSRPグループのプライオリティ値は何ですか?
A. 0
B. 1
C. 2
D. 3
Answer: D
NEW QUESTION: 2
What are the components on which reporting is based upon, in profit center accounting?
There are 2 correct answers to this question.
A. Cost elements
B. Cost centers
C. Profit centers
D. Profit center hierarchy
Answer: C,D
NEW QUESTION: 3
セキュリティエンジニアは、管理者からの依頼により、設定ボードの新しいメンバーになりました。同社は今年から2つの新しい主要なITプロジェクトを開始しており、アプリケーション展開のセキュリティを計画したいと考えています。理事会は主に、アプリケーションが連邦政府の査定および承認基準に準拠しているかどうかを懸念します。セキュリティエンジニアは、両方のアプリケーションのセキュリティ評価をいつ行うべきかを決定するためのタイムラインを求め、その後のコンフィギュレーションボード会議には出席しません。セキュリティエンジニアがセキュリティアセスメントを実行するだけの場合は、システム承認の次の手順のうちセキュリティエンジニアを省略したのはどれですか。
A. セキュリティ管理基準を確立します
B. ユーザー受け入れテストの結果を確認します
C. ソフトウェア開発セキュリティ標準に従ってアプリケーションをビルドします
D. 利害関係者と相談して、どの基準を省略できるかを判断します。
Answer: A
Explanation:
A security baseline is the minimum level of security that a system, network, or device must adhere to. It is the initial point of reference for security and the document against which assessments would be done.
Incorrect Answers:
B: Building the application with secure coding is the programmers' duty.
C: User acceptance testing is part of the development process
D: Standards are not security concerns.
References:
Gregg, Michael, and Billy Haines, CASP CompTIA Advanced Security Practitioner Study Guide, John Wiley & Sons, Indianapolis, 2012, pp. 272-273
NEW QUESTION: 4
Refer to the exhibit.
The network is converged.After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and
208.149.23.96?
A. 208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0
2 08.149.23.96[110/3] via 190.173.23.10, 00:00:16, Serial1/0
B. 208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0
2 08.149.23.96[110/3] via 190.173.23.10, 00:00:16, FastEthemet0/0
C. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0
2 08.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
D. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, Serial1/0
208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
Answer: C
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 /
Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets ->
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps =
64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link ->
Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -> they are surely incorrect.