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Helen [10]
1 year ago
11

Who benefits the most from billing by the second for cloud resources, such as virtual machines?.

Computers and Technology
1 answer:
scoundrel [369]1 year ago
7 0

Clients that run numerous virtual machines.

<h3>What is a virtual machine?</h3>
  • A virtual machine is the virtualization or emulation of a computer system in computing.
  • The functionality of a physical computer is provided by virtual machines, which are built on computer architectures.
  • Their use may necessitate specialized hardware, software, or a combination of both.
  • Virtual machines' primary function is the simultaneous operation of several operating systems on a single piece of hardware.
  • Without virtualization, running two different physical units would be necessary to run different operating systems, such as Windows and Linux.
  • Through the use of virtualization technologies, virtual machines are made possible.
  • Multiple virtual machines (VMs) can run on a single machine thanks to virtualization, which simulates virtual hardware using software.
  • The real machine is referred to as the host, and any virtual machines running on it as the guests.

To learn more about virtual machines, refer to:

brainly.com/question/27961159

#SPJ4

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What kind of physical device is an evil twin access point? What does the evil twin do after initial association when the victim
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A PC such as a notebook computer.The evil twin gets the encrypted frame from the affected host,A VPN  encrypts a  frame with its own VPN key.

Explanation:

The evil twin after initial association when the victim client will establish a secure wireless connection with the victim client. The connection uses a key client-EF for encryption such that when the host transmits an encrypted frame it is transmitted to the evil twin. A VPN is an encrypted virtual private network used to access networks that are not trusted. The client encrypts a frame with a VPN key (Key Client-Server) which it shares with the server. The frame is further encrypted by the key it shares with the evil twin (Victim Client-ET) The evil twin will then receive a double encrypted frame. However, it will only be able to decrypt the Victim Client-ET key but not the VPN key hence it will not be able to read the message sent through the frame.

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Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
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Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)    

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

Explanation:  

Unification: Any substitution that makes two expressions equal is called a unifier.  

a) P(B,A,B), P(x,y,z)  

Use { x/B}  

=> P(B,A,B) , P(B,y,z)  

Now use {y/A}  

=> P(B,A,B) , P(B,A,z)  

Now, use {z/B}  

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }  

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q  

c. Older(Father(y),y), Older(Father(x),John)  

Use {y/x}  

=> Older(Father(x),x), Older(Father(x),John)  

Now use { x/John }  

=> Older(Father(John), John), Older(Father(John), John)  

Thus , mgu ={ y/x , x/John }  

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))  

Use { y/x }  

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}  

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B  

e) P(f(x), x, g(x)), P(f(y), A, z)  

Use {x/y}  

=> P(f(y), y, g(y)), P(f(y), A, z)  

Now use {z/g(y)}  

P(f(y), y, g(y)), P(f(y), A, g(y))  

Now use {y/A}  

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

7 0
3 years ago
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