Authentic Fortinet Certification Exam FCSS_NST_SE-7.6 Questions Answers

The correct answers are A and D .

The Network Security Support Engineer 7.6 Study Guide explains that automation stitches consist of a trigger and one or more actions , and that they can detect events such as high CPU and conserve mode across the Security Fabric

The FortiOS administration guide then gives the exact practical example for A :

“Automation stitches can be created to run a CLI script and send an email message when memory or CPU usage exceeds specified thresholds.”

It also shows examples where the email body contains the script output using:

%%results%%

That directly confirms A .

For D , the FortiOS administration guide states:

“The URI and HTTP body can use parameters from logs or previous action results.”

It also explains the execution model:

“The stitch Action execution can be set to either Sequential or Parallel. In sequential execution actions will execute one after another with a delay (if specified). … In parallel execution all actions will execute immediately when the stitch is triggered.”

A concrete example shows a second action using the output of the first action:

“This string for the body text includes the results from the preceding CLI script action.” with

Body=%%results%%...

That confirms D .

Why the other options are wrong:

B is wrong because automation stitches can be triggered by IPS events, but the documentation does not describe them as modifying packet headers or payloads. Instead, they perform actions such as email, CLI script, webhook, quarantine, and similar automated responses

C is wrong because delay is associated with sequential execution, not parallel execution. The guide explicitly says: “A delay can be added before an action if Sequential action execution is used.”

So the verified answers are: A, D .

The study guide states:

“The kernel memory slabs are collections of objects with a common purpose. The kernel uses them to store information in memory.”

It also gives the exact calculation rule:

“Total slab size = available objects x object size”

From the exhibit:

tcp_session 3 5 1500 ...

So:

available objects = 5

object size = 1500

Therefore:

Total slab size = 5 × 1500 = 7500 kB

That makes D correct, and it is associated with the kernel , not user space.

Why the other options are wrong:

A is wrong because sctp_session 0 0 1600 ... gives 0 × 1600 = 0 , but slabs are associated with the kernel , not user space.

B is wrong because ip_session 1 3 1200 ... gives 3 × 1200 = 3600 , but again slabs are kernel memory, not user space.

C is wrong because ip6_session 0 0 1300 ... gives 0 × 1300 = 0 , not 1300.