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NEW QUESTION: 1
ある企業は、すべてのVMを予約済みインスタンスに移動することを検討しています。これにより、各インスタンスで20%節約できます。会社が移動するインスタンスを以下に示します。

これらすべてのVMをリザーブドインスタンスに変換することで、会社が毎年節約できる金額は次のうちどれですか?
A. 53280ドル
B. $ 3840
C. 5550ドル
D. 1600ドル
E. $ 13320
F. $ 1110
Answer: E

NEW QUESTION: 2
Refer to the exhibit.

An administrator examines the configuration of an AP from apboot mode shown in the exhibit.
What can the administrator determine from the configuration?
A. AP terminated on a Mobility Controller in a cluster.
B. RAP terminated on a stand-alone controller.
C. AP terminated on a non-cluster Mobility Controller.
D. RAP terminated on a Mobility Master.
Answer: A

NEW QUESTION: 3
Ihr Unternehmen hat Büros in Montreal und Seattle.
Das Netzwerk enthält eine Active Directory-Gesamtstruktur. Die Gesamtstruktur enthält eine Domäne und zwei Standorte mit den Namen Montreal und Seattle. Das Büro in Seattle enthält zwei Domänencontroller mit den Namen DC4 und DC5.
Sie verlegen DC5 physisch vom Büro in Seattle in das Büro in Montreal.
Sie stellen fest, dass DC5 weiterhin Benutzer im Büro in Seattle authentifiziert.
Sie müssen sicherstellen, dass die Benutzer des Büros in Seattle von DC4 authentifiziert werden, es sei denn, DC4 ist nicht verfügbar.
Was solltest du zuerst tun?
A. Ändern Sie unter Active Directory-Standorte und -Dienste die NTDS-Standorteinstellungen für das Büro in Seattle.
B. Verschieben Sie unter Active Directory-Benutzer und -Computer das Computerkonto von DC5.
C. Ändern Sie unter Active Directory-Benutzer und -Computer die NTDS-Einstellungen für das Computerkonto von DC5.
D. Verschieben Sie das DC5-Serverobjekt von Active Directory-Standorten und -Diensten.
Answer: D
Explanation:
Erläuterung
Verweise:
https://blogs.technet.microsoft.com/canitpro/2015/03/03/step-by-step-setting-up-active-directory-sites-subnets-si

NEW QUESTION: 4
An administrator is working with the 192.168.4.0 network, which has been subnetted with a /26 mask. Which two
addresses can be assigned to hosts within the same subnet? (Choose two.)
A. 192.168.4.125
B. 192.168.4.128
C. 192.168.4.132
D. 192.168.4.63
E. 192.168.4.67
F. 192.168.4.61
Answer: A,E
Explanation:
Increment: 64 (/26 = 11111111.11111111.11111111.11000000)
The IP 192.168.4.0 belongs to class C.
The default subnet mask of class C is /24 and it has been subnetted with a /26
mask so we have 2(26-24) = 22 = 4 sub-networks:
1st subnet: 192.168.4.0 (to 192.168.4.63)
2nd subnet: 192.168.4.64 (to 192.168.4.127)
3rd subnet: 192.168.4.128 (to 192.168.4.191)
4th subnet: 192.168.4.192 (to 192.168.4.225)
In all the answers above, only answer C and D are in the same subnet.
Therefore only IPs in this range can be assigned to hosts.