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tigry1 [53]
3 years ago
13

Who sends the Certificate, the client, the server, or both? A certificate is sent by one party to let the other party authentica

te that it is who it claims to be. Based on this usage, you should be able to guess who sends the certificate and check the messages in your trace.
Computers and Technology
1 answer:
ohaa [14]3 years ago
3 0

Answer: Both the server and the client send certificates

Explanation:

In a given network session the server sends a certificate to the client . The client is able to verify that the server's identity, the authentication uses a public encryption key and a digital signature .The client in turn retrieves a private key they can use to create a digital signature . A certificate is then sent by the client across the network to the server and it uses it to authenticate the client's identity . Once the server has confirmed that the client is who they say they are it grants access for the verified identification.

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Counting calculus students. About A university offers 3 calculus classes: Math 2A, 2B and 2C. In both parts, you are given data
OleMash [197]

Answer:

Part (a) n(A∩B∩C) = 4

Part (b) n(A∪B∪C) = 54

Explanation:

n(A) = no. of students who took Math 2A

n(B) = no. of students who took Math 2B

n(C) = no. of students who took Math 2C

n(A∩B) = no. of students who took both Math 2A and 2B

n(A∩C) = no. of students who took both Math 2A and 2C

n(B∩C) = no. of students who took both Math 2B and 2C

n(A∩B∩C) = no. of students who took all three Math 2A, 2B and 2C

n(A∪B∪C) = no. of total students in a group

∩ represents Intersection and ∪ represents Union

Part (a)

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)  

Where n(A∩B∩C) represents the number of students who took all three classes and n(A∪B∪C) represents the total number of students in group A

157 = 51 + 80 + 70 - 15 - 20 - 13 + n(A∩B∩C)

Re-arranging the equation to solve for n(A∩B∩C) since we want to find out those students who took all three classes

n(A∩B∩C) = 157 - 51 - 80 - 70 + 15 + 20 + 13

n(A∩B∩C) = 4

So there are 4 students in group A who took all three classes

Part (b)

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)

This time we are given n(A∩B∩C) students who took all three classes and want to find n(A∪B∪C) that is total number of students

n(A∪B∪C) = 28 + 28 + 25 - 11 - 9 - 10 + 3

n(A∪B∪C) = 54

So there are total 54 students in group B

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