Answer:
Yes, they can be aggregated into a single network address of 88.62.0.0/16
Explanation:
Since they all will be using the same outgoing interface so they can be grouped according to next hop which appears to be the same for all four IP address.
First convert all the network addresses into binary.
DECIMAL BINARY CLASS OF IP
88.62.108.0/21 01011000.00111110.01101100.00000000 Class A
88.62.116.0/21 01011000.00111110.01110100.00000000 Class A
88.62.120.0/21 01011000.00111110.01111000.00000000 Class A
88.62.128.0/21 01011000.00111110.10000000.00000000 class A
Looking at the binary form of the address from left to right it can be observed that all for address are identical in the first two octet, that is up to the 16th digit.
So the new subnet mask will be /16, using any of the four address and the /16 mask will can obtain the new network address
All four address can be summarized as 88.62.0.0/16
Answer:
B
Explanation:
selection sort for each pass sorts one element in right place . so we can stop it after 5 passes as we need only 5 heaviest cows info.
Answer is B
It is letter d
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