The solution is 256-248 = 8, 16, and 24. The broadcast address of the 16 subnet, where this host is located, is 23, and the range of acceptable hosts is 17–22.
This equals 254 after subtracting the 2 reserved addresses. You will therefore receive 254 valid hosts with the chosen subnet mask. The ability to assign 62 hosts an IP Address via a /24 subnet would be possible regardless of whether the IP Address is private or public. /16 would produce 65,534 hosts (or 124.125.0.1 - 124.125.1) as a result. This indicates that a "/16" leaves the final 16 bits (or final two integers) available for usage in specified addresses and a "/8" leaves the final 24 bits available for use.
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