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HPE3-U01 Exam - Topic 1 Question 26 Discussion

How many addresses does the 255.255.252.0 provide?
C) 8192
A) 16384
B) 32768
D) 1024

HPE3-U01 Exam - Topic 1 Question 26 Discussion

Actual exam question for HP's HPE3-U01 exam
Question #: 26
Topic #: 1
[All HPE3-U01 Questions]

How many addresses does the 255.255.252.0 provide?

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Suggested Answer: C, E

by Claribel at Oct 18, 2023, 09:22 PM

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Erick
7 months ago
1024 seems low for a subnet mask like that.
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Andra
7 months ago
I disagree, I thought it was 8192.
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Della
7 months ago
Wait, really? I thought it was more than that.
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Willard
7 months ago
Definitely option D, 1024 is correct!
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Lynelle
7 months ago
It provides 1024 addresses.
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Tamra
8 months ago
I thought 255.255.252.0 gives a larger range, like 4096 total addresses, but I can't recall the exact breakdown.
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Quentin
8 months ago
I feel like I might be mixing up the subnet masks. Wasn't there a question about /22 giving 1024 usable addresses?
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Kelvin
8 months ago
I remember practicing similar questions, and I think the formula is 2^(32 - subnet bits). So, maybe it's 1024 addresses?
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Dorthy
8 months ago
I think 255.255.252.0 is a /22 subnet, but I'm not entirely sure how to calculate the total addresses from that.
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Justine
8 months ago
I got this! The subnet mask 255.255.252.0 means the last 2 bits are available for host addresses. That gives us 2^2 = 4 subnets, and each subnet has 2^6 = 64 addresses. So the total number of addresses is 4 * 64 = 8192. Easy peasy!
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Leontine
8 months ago
Okay, let's see. The subnet mask 255.255.252.0 means we have 2 bits available in the last octet, so that should give us 2^2 = 4 subnets. I think the answer is 8192, but I'll double-check my work.
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Misty
8 months ago
Hmm, I'm a bit unsure about this one. I know subnet masks are important, but I'm not totally confident in my ability to work through these types of problems quickly. I'll need to think it through carefully.
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Vilma
8 months ago
This looks like a straightforward subnet mask question. I'll need to calculate the number of available addresses based on the given subnet mask.
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Arthur
8 months ago
I think Cisco CloudLock is the best option here since it provides visibility and protection against data exfiltration, which are the key requirements mentioned in the question.
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Annmarie
8 months ago
Hmm, I'm a little unsure about the details here. I know switches are better than hubs, but I'll need to carefully review the options to make sure I select the right benefits.
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Aracelis
8 months ago
This seems straightforward. I'd go with creating a document set - that's the most efficient way to print multiple documents at once.
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Gail
8 months ago
I'm pretty sure the Cost to Income Ratio will deteriorate because higher salaries mean increased costs.
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Vernice
1 year ago
I'm just going to guess and hope for the best. Networking isn't my strongest suit, but I'll give it a shot.
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Bettina
1 year ago
Wait, is this a trick question? I'm pretty sure the answer is 1024, but you never know with these exams.
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Svetlana
1 year ago
Ah, I remember this from the practice questions. The answer is C) 8192. Gotta love those subnet masks!
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Felicidad
11 months ago
Definitely, practice makes perfect!
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Lucy
12 months ago
So true, but once you practice enough, it becomes easier to remember.
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Aleta
12 months ago
Yes, you're right! Subnet masks can be tricky.
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Tiara
1 year ago
I think the answer is C) 8192.
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Emelda
1 year ago
I remember learning about this in my networking class. Let me see, 255.255.252.0 should give us 1024 addresses, right?
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Avery
11 months ago
Yes, that's correct. It gives 8192 addresses.
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Lizbeth
12 months ago
Oh really? I must have remembered it wrong. Thanks for correcting me.
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Candida
12 months ago
No, it actually provides 8192 addresses.
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Tresa
1 year ago
Hmm, I see your point. Let's calculate it together to be sure.
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Antione
1 year ago
Hmm, this is a tricky one. I'm going to have to think about this subnet mask carefully.
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Genevieve
1 year ago
B) 32768
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Lynelle
1 year ago
A) 16384
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Alishia
1 year ago
I disagree, I believe the answer is B) 32768.
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Tresa
1 year ago
I think the answer is A) 16384.
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