Valid 700-805 Braindumps & Cisco 700-805 Test Book - 700-805 Free Sample Questions - FreeTorrent
Exam Code: 700-805
Exam Name: Cisco Renewals Manager
Version: V22.75
Q & A: 580 Questions and Answers
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NEW QUESTION: 1
You have two Skype for Business Server 2015 Front End servers.
You plan to install a third Front End server. User1 must be able to install a new Skype for Business Server 2015 Front
You need to delegate the minimum level of Skype for Business Server permissions to User1.
Which two actions should you perform? Each correct answer presents part of the solution. Choose two.
A. Grant User1 membership to the RTCUniversalServerAdmins group.
B. Assign User1 the CsServerAdministrator role.
C. Assign User1 the CsLocationAdministrator role.
D. Assign User1 the CsLocationAdministrator role.
E. From Skype for Business Management Shell, run the following cmdlet: Grant-CsSetupPermission -ComputerOU "ou
= CsServers,dc = litwareinc,dc=com
F. From Skype for Business Management Shell, run the following cmdlet: Update-CsAdminRole
G. Grant User1 membership to the RTCUniversalUserAdmins group.
Answer: E,F
NEW QUESTION: 2
質問に答えるために、以下の情報を参照してください。
大規模な組織では、一意の識別子を使用して、すべてのシステム・セッションの開始時に、それらを必要とします。 アプリケーションへのアクセスはジョブの分類に基づいています。 組織はアクセス制御と違反の定期的な独立したレビューの対象となります。 組織はまた、選択された情報とプロセスのためのブランチオフィスとの安全なバックアップとリカバリ戦略への安全な接続を使用しています。
アクセス制御ログは識別子のほかには何が含まれなければなりませんか。
A. 拒否されたアクセスの試み
B. アクセスの時間
C. 関連クリアランス
D. 秘密区分
Answer: B
NEW QUESTION: 3
Inventory Record is one of the major input to the
A. Capacity Requirements planning (CRP)
B. Manufacturing Planning and Control (MPC) Function
C. Production activity Control
D. Material Requirements planning (MRP) System
Answer: D
NEW QUESTION: 4
BeigeBox, a computer manufacturer, can construct computer systems with any one of 4 processors, 2 memory configurations, 3 hard drives, 2 monitors, 2 keyboards, and 2 CD drives. How many unique configurations of computer systems can BeigeBox construct?
A. 0
B. 1
C. 2
Answer: B
Explanation:
The multiplication rule of counting states that the number of combinations available when there are n_1 options in one aspect, n_2 in another, and so on, up to n_k, is n_1 * n_2 * ... * n_k. In this case, the number of combinations is 4*2*3*2*2*2 = 192.
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