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NEW QUESTION: 1
Your company's network has client computers that run Windows 7.
A software vendor releases version 2 of an application that your company uses.
Your company currently uses version 1. Version 1 and version 2 are not compatible.
You plan to deploy version 2 of the application.
You have the following requirements:
Users must be able to run both versions of the application on their computers.
Version 2 must be available when a client computer is not connected to the network.
You need to plan a software deployment process that meets the requirements.
What should you do?
A. Deploy version 2 of the application by using a Microsoft System Center Configuration Manager package.
B. Deploy version 2 of the application by using a Group Policy Software Installation policy.
C. Deploy version 2 of the application as a Remote Desktop Services RemoteApp.
D. Deploy version 2 of the application by using Microsoft Application Virtualization (App-V).
Answer: D
Explanation:
Microsoft Application Virtualization (MS App-V) platform allows applications to be deployed in real- time to any client from a virtual application server. It removes the need for local installation of the applications. Instead, only the App-v client needs to be installed on the client machines. All application data is permanently stored on the virtual application server. Whichever software is needed is either streamed or locally cached from the application server on demand and run locally. The App-V stack sandboxes the execution environment so that the application does not make changes to the client itself (OS File System and/or Registry). App-V applications are also sandboxed from each other, so that different versions of the same application can be run under App-V concurrently.
http://en.wikipedia.org/wiki/Microsoft_App-V

NEW QUESTION: 2

A. JBOD
B. Distributed RAID
C. Traditional RAID
D. Drive mirroring
Answer: B

NEW QUESTION: 3
A normal distribution has a mean of 39 and a variance of 625. The area under the distribution between 50 and 75 equals ________.
A. 0.745
B. 0.255
C. 0.212
D. 0.319
Answer: B
Explanation:
Explanation/Reference:
Explanation:
First note that the area between two numbers under a probability distribution is the same as the probability of getting a number between those two numbers. The standard deviation of the distribution equals sqrt (625) = 25. The z-value of 50 equals (50-39)/25 = 0.44. The z-value of 75 equals (75-39)/25 = 1.44. Using the Normal probability tables, we get P(z < 0.44) = 0.67 and P(z < 1.44) = 0.9251. The area between 50 and 75 is then given by P(0.44 < z < 1.44) = P(z < 1.44) - P(z < 0.44) = 0.9251 - 0.67 = 0.255.

NEW QUESTION: 4

A. FLOGI
B. PRLI
C. PLOGI
D. WWPN
Answer: A