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NEW QUESTION: 1
A PC cannot connect to the network. A network card was purchased without documentation or driver discs. Which of the following is the BEST way to install it into this machine?
A. Copy drivers to flash drive and install
B. Purchase the disc online and install
C. Download and install the drivers
D. Run Windows update to install drivers
Answer: A
Explanation:
Topic 3, Volume C

NEW QUESTION: 2

A. Option A
B. Option D
C. Option C
D. Option B
Answer: D

NEW QUESTION: 3
Ihr Netzwerk enthält eine Active Directory-Gesamtstruktur mit dem Namen contoso.com.
Sie müssen der Gesamtstruktur eine neue Domain mit dem Namen fabrikam.com hinzufügen.
Welchen Befehl sollten Sie ausführen? Wählen Sie zum Beantworten die entsprechenden Optionen im Antwortbereich aus.

Answer:
Explanation:

Explanation

References:
https://technet.microsoft.com/en-us/library/hh974722(v=wps.630).aspx

NEW QUESTION: 4
質問をドラッグアンドドロップ
IPv4ネットワークサブネットを左側から右側の正しい使用可能なホスト範囲にドラッグアンドドロップします

Answer:
Explanation:

Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet 172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111
1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is
172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or
172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet.
Therefore the last usable host is 172.28.229.254.

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