Based on the configuration written below, the most likely cause of the problem is Incorrect subnet mask.
<h3>What is an Incorrect Subnet Mask?</h3>
The issue of an Incorrect Subnet Mask will take place if a network uses a subnet mask that is not theirs for its address class, and a client is still said to be configured with the same default subnet mask for the address class, and thus communication tend to fail to some closeby networks.
Therefore, Based on the configuration written below, the most likely cause of the problem is Incorrect subnet mask.
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See full question below
You manage a network that has multiple internal subnets. You connect a workstation to the 192.168.1.0/24 subnet.
This workstation can communicate with some hosts on the private network, but not with other hosts. You run ipconfig /all and see the following:
Ethernet adapter Local Area Connection: Connection-specific DNS Suffix . : mydomain.local Description . . . . . . . : Broadcom network adapter Physical Address. . . . . . : 00-AA-BB-CC-74-EF DHCP Enabled . . . . . . . : No Autoconfiguration Enabled. . . : Yes IPv4 Address . . . . . . . : 192.168.1.102(Preferred) Subnet Mask. . . . . . . . : 255.255.0.0 Default Gateway . . . . . . : 192.168.1.1 DNS Servers . . . . . . . : 192.168.1.20 192.168.1.27
What is the most likely cause of the problem?
Answer:
Explanation:
There are all sorts of possibilities for, say, inserting new technologies into existing processes. But most of these improvements are incremental. They are worth doing; in fact, they may be necessary for survival. No self-respecting airline, for instance, could do without an application that lets you download your boarding pass to your mobile telephone. It saves paper, can't get lost and customers want it.
But while it's essential to offer applications like the electronic boarding pass, those will not distinguish a company. Electronic boarding passes have already been replicated by nearly every airline. In fact, we've already forgotten who was first.
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Answer:
numbers = 1:1:100;
for num=numbers
remainder3 = rem(num,3);
remainder5 = rem(num,5);
if remainder3==0
disp("Yee")
else
if remainder3 == 0 && remainder5 == 0
disp ("Yee-Haw")
else
if remainder5==0
disp("Haw")
else
disp("Not a multiple of 5 or 4")
end
end
end
end
Explanation:
- Initialize the numbers variable from 1 to 100.
- Loop through the all the numbers and find their remainders.
- Check if a number is multiple of 5, 3 or both and display the message accordingly.
World Wide Web (WWW)? I honestly don't know.