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NEW QUESTION: 1

A. Option B
B. Option A
C. Option C
D. Option D
Answer: D
Explanation:
http://technet.microsoft.com/en-us/library/cc732757.aspx
Pnputil.exe is a command line utility that you can use to manage the driver store. You can use Pnputil to add driver packages, remove driver packages, and list driver packages that are in the store. http://technet.microsoft.com/en-us/library/ff800798(v=ws.10).aspx http://technet.microsoft.com/en-us/library/hh825072.aspx http://technet.microsoft.com/en-us/library/hh825251.aspx

NEW QUESTION: 2
You have several virtual machines (VMs) that run in Azure. You also have a single System Center 2012 R2 Configuration Manager (SCCM) primary site on-premises.
You have the following requirements:
- All VMs must run on the same virtual network.
- Network traffic must be minimized between the on-premises datacenter and Azure.
- The solution minimize complexity.
You need to use SCCM to collect inventory and deploy software to Azure VMs.
What should you do first?
A. Enable and configure Operations Insights in Azure.
B. Configure client push for the Azure virtual network.
C. Install a cloud distribution point on an Azure VM.
D. Install a secondary site underneath the primary site onto an Azure VM.
Answer: C
Explanation:
Ref: http://www.aidanfinn.com/?p=16047

NEW QUESTION: 3
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.255.255.255.254
B. 許可10.0.0.1.0.0.0.254
C. 許可10.0.0.0.0.0.0.1
D. 許可10.0.0.1.0.0.0.0
Answer: B
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network .x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.