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Annette [7]
4 years ago
14

5255555555555+55555555555555/1111*99442

Computers and Technology
2 answers:
Marina CMI [18]4 years ago
7 0

Answer:

(502763134343429265 /101)

Explanation:

bixtya [17]4 years ago
7 0

Answer :what?

Explanation:

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Read 2 more answers
) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in
erik [133]

Answer:

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28

4 0
3 years ago
2.Consider the following algorithm and A is a 2-D array of size ???? × ????: int any_equal(int n, int A[][]) { int i, j, k, m; f
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Answer:

(a) What is the best case time complexity of the algorithm (assuming n > 1)?

Answer: O(1)

(b) What is the worst case time complexity of the algorithm?

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(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1) .

(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).

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