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Exam Code: XK0-005
Exam Name: CompTIA Linux+ Certification Exam
Version: V22.75
Q & A: 580 Questions and Answers
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NEW QUESTION: 1
사용자가 Auto Scaling 그룹에서 조정 프로세스를 일시 중지했습니다. 인스턴스 수를 늘리기 위한 조정 활동이 이미 진행 중입니다.
정학은 그 활동에 어떤 영향을 미칩니 까?
A. 인스턴스를 일시적으로 중지
B. 인스턴스 시작을 일시 중지하고 Auto Scaling이 재개된 후에만 시작
C. 효과가 없습니다. 스케일링 활동 계속
D. 인스턴스를 종료합니다
Answer: C
Explanation:
The user may want to stop the automated scaling processes on the Auto Scaling groups either to perform manual operations or during emergency situations. To perform this, the user can suspend one or more scaling processes at any time. When this process is suspended, Auto Scaling creates no new scaling activities for that group. Scaling activities that were already in progress before the group was suspended continue until completed.
http://docs.aws.amazon.com/AutoScaling/latest/DeveloperGuide/AS_Concepts.html
NEW QUESTION: 2
A mule application is being designed to perform product orchestration. The Mule application needs to join together the responses from an inventory API and a Product Sales History API with the least latency.
To minimize the overall latency. What is the most idiomatic (used for its intended purpose) design to call each API request in the Mule application?
A. Call each API request in a separate Async scope
B. Call each API request in a separate route of a Parallel For Each scope
C. Call each API request in a separate route of a Scatter-Gather
D. Call each API request in a separate lookup call from Dataweave reduce operator
Answer: C
NEW QUESTION: 3
What is the default behavior of NAT control on Cisco ASA Software Version 8.3?
A. It will prevent traffic from traversing from one enclave to the next without proper access configuration.
B. It will allow traffic to traverse from one enclave to the next without proper access configuration.
C. It will deny all traffic.
D. NAT control has been deprecated on Cisco ASA Software Version 8.3.
Answer: D
NEW QUESTION: 4
How many times will the function counter() be executed in the following code?
function counter($start, &$stop)
{
if ($stop> $start)
{
return;
}
counter($start--, ++$stop);
$ start= 5;
$ stop= 2;
counter($start, $stop);
A. 0
B. 1
C. 2
D. 3
Answer: C
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