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Microsoft AZ-104 Exam - Topic 4 Question 70 Discussion

Posted by Glory on Dec 11, 2022 at 12:46 PM

You have an on-premises network.

You have an Azure subscription that contains three virtual networks named VNET1, VNET2, and VNET3. The virtual networks are peered and connected to the on-premises network. The subscription contains the virtual machines shown in the following table.

You need to monitor connectivity between the virtual machines and the on-premises network by using Connection Monitor. What is the minimum number of connection monitors you should deploy?

Suggested Answer: B

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11 comments · newest first
CA
Catalina 11 months ago
Not sure about that, seems too simple for such a complex network.
DO
Donette 11 months ago
Definitely 1, it can monitor multiple VMs at once.
MA
Maybelle 12 months ago
Wait, really? I thought we needed one for each VM.
TA
Tamar 12 months ago
I think 2 would be safer, just to cover all bases.
LE
Leigha 1 year ago
You only need 1 connection monitor for this setup.
AN
An 1 year ago
I'm a bit confused; could it be just one monitor if they are all peered? I need to double-check that concept.
LI
Lindsey 1 year ago
I practiced a similar question, and I think the answer might be three because we need to cover all the VMs effectively.
RO
Rosendo 1 year ago
I think we might need at least two monitors since we have multiple VMs across different networks.
CU
Curtis 1 year ago
I remember something about needing one connection monitor per virtual network, but I'm not sure if that's the case here.
KI
Kimberely 1 year ago
Hmm, this is a tricky one. I'm not entirely sure which option to choose here. I know expressions can be used to evaluate conditions, but I'm not confident that's the right answer. I might need to review the Power Automate documentation to be sure.
LO
Lorriane 1 year ago
I've got this! A HIPS is all about monitoring and protecting the individual host, not the network. So it can't detect events scattered across the network, making statement A false. And it's not related to IP address sharing, so statement B is also false.
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