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HPE3-U01 Exam - Topic 2 Question 54 Discussion

Actual exam question for HP's HPE3-U01 exam
Question #: 54
Topic #: 2
[All HPE3-U01 Questions]

Given the 172.16.0.0 255.255.255.0 IP segment, how many IP addresses can be assigned to host and network devices?

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Suggested Answer: A

by Abel at Dec 04, 2024, 01:28 PM

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Raylene
3 months ago
I thought it was 126 too, but I guess I was wrong!
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Lettie
3 months ago
128 is way off, it's definitely more than that.
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Domonique
3 months ago
Wait, are we sure about that?
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Josephine
4 months ago
Totally agree, 254 is the right number.
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Valentin
4 months ago
It's 254 usable IPs, not 126!
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Louvenia
4 months ago
I thought it was 128 usable addresses, but now I'm confused about the network and broadcast addresses. Can someone clarify that?
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Galen
4 months ago
I practiced a question like this, and I think the answer is 126 usable IPs. The formula is 2^n - 2, where n is the number of host bits.
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Raylene
4 months ago
I'm not entirely sure, but I remember something about needing to calculate usable addresses. Is it 128 or 126? I feel like I've seen similar questions before.
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Billi
5 months ago
I think the subnet mask 255.255.255.0 gives us a /24, which means we have 256 total addresses, but we need to subtract 2 for the network and broadcast addresses, right?
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Kindra
5 months ago
I've got this one! The subnet mask 255.255.255.0 means we have 8 bits for the host portion. 2^8 = 256 possible host addresses, but we need to subtract the network address (172.16.0.0) and the broadcast address (172.16.0.255), so the answer is 254 available IP addresses.
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Owen
5 months ago
This looks straightforward. The subnet mask is 255.255.255.0, so the network has 256 possible IP addresses (2^8). Subtracting the network and broadcast addresses, we get 254 available IP addresses.
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Aliza
5 months ago
I'm a bit confused on how to calculate the number of available IP addresses. Can someone walk me through the process?
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Annalee
5 months ago
Okay, let's think this through step-by-step. The IP segment is 172.16.0.0 with a subnet mask of 255.255.255.0, so the network portion is 172.16.0.0 and the host portion is 0.0.0.0. The number of host bits is 8, so the number of possible host addresses is 2^8 - 2 = 256 - 2 = 254.
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Jodi
9 months ago
I'm just hoping the answer isn't E) 42. That would be way too cosmic for a networking exam!
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Louis
8 months ago
I agree, C) 128 sounds right for that IP segment.
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Dean
8 months ago
I hope it's C) 128. That seems like a reasonable number.
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Tamie
8 months ago
C) 128
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Rebeca
8 months ago
A) 126
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Carlota
10 months ago
I'm with Jeannetta on this one. A) 126 has got to be the correct answer. Who would waste two IP addresses for the network and broadcast addresses? Silly question.
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Jaime
10 months ago
Woah, this is a tricky one. Let me think... B) 6 seems way too low, and D) 4 just doesn't make sense. I'm going with C) 128, it feels right.
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Micah
8 months ago
Yes, C) 128 is the correct answer for the number of IP addresses that can be assigned.
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Jolanda
9 months ago
I agree, C) 128 sounds right to me too.
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Anastacia
9 months ago
I think C) 128 is the correct answer.
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Jeannetta
10 months ago
I'm not so sure about that. Doesn't the network address and broadcast address take up two of those addresses? I'm leaning towards A) 126.
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Ligia
9 months ago
Great, glad we figured that out together!
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Mona
10 months ago
I agree, the correct answer is A) 126.
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Earlean
10 months ago
So that leaves us with 126 usable IP addresses for host and network devices.
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Ashton
10 months ago
I think you're right, the network and broadcast addresses do take up two of the addresses.
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Remona
10 months ago
Hmm, I think the answer is C) 128. That's the number of available host addresses in a /24 subnet, right?
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Janessa
10 months ago
So that leaves us with 254 usable IP addresses for host devices. The answer should be C) 128.
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Isabella
10 months ago
I agree with Tuyet. With a /24 subnet mask, there are 256 total IP addresses, but 2 are reserved for network and broadcast addresses.
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Tuyet
11 months ago
I think the answer is C) 128.
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