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NEW QUESTION: 1
Review the boot environment information displayed on your system:
Which two options accurately describe the newBE boot environment?
A. You can create a snapshot of it.
B. It is activated but unbootable.
C. It cannot be activated.
D. It has been deleted and will be removed at the next reboot.
E. It cannot be destroyed.
F. It cannot be renamed.
Answer: C,F
Explanation:
Explanation/Reference:
Explanation:
If the boot environment is unbootable, it is marked with an exclamation point (!) in the Active column in the beadm list output.
The beadm command restricts actions on unbootable boot environments as follows:
You cannot activate an unbootable boot environment. (B)
You cannot destroy a boot environment that is both unbootable and marked as active on reboot.
You cannot create a snapshot of an unbootable boot environment.
You cannot use an unbootable boot environment or boot environment snapshot with the -e option of beadm create.
You cannot rename an unbootable boot environment. (C)
NEW QUESTION: 2
다음 표에 나와 있는 리소스가 포함된 Azure 구독이 있습니다.
NIC1이라는 네트워크 인터페이스를 만들어야 합니다.
어느 위치에서 NIC1을 만들수 있습니까?
A. 미국 동부만 해당.
B. 미국 동부 및 북유럽에만 해당됩니다.
C. 미국 동부 및 서유럽에만 해당됩니다.
D. 미국 동부, 서유럽 및 북유럽.
Answer: A
Explanation:
설명
NIC를 만들 때 가상 네트워크가 필요합니다. 네트워크 인터페이스에 대한 가상 네트워크를 선택하십시오. 네트워크 인터페이스와 동일한 가입 및 위치에 있는 가상 네트워크에만 네트워크 인터페이스를 할당할수 있습니다. 네트워크 인터페이스가 생성되면 할당된 가상 네트워크를 변경할 수 없습니다.
네트워크 인터페이스를 추가 한 가상 머신도 네트워크 인터페이스와 동일한 위치 및 구독에 있어야 합니다.
참고 문헌 :
https://docs.microsoft.com/en-us/azure/virtual-network/virtual-network-network-interface
NEW QUESTION: 3
Ein Projektteam vervollständigt die Identifizierung der Projekt- und Produktanforderungen. Welche der Erneuerungen wurden verwendet, um dies zu erreichen?
A. Brainstorming, Affinitätsdiagramme und Mind Mapping
B. Benchmarking, Netzwerkdiagramm und Trendanalyse
C. Affinitätsdiagramme, Benchmarking und Beobachtung / Konversation
D. Projektstrukturplan (PSP), Trendanalyse und Beobachtung / Konversation
Answer: A
NEW QUESTION: 4
Given:
public class Emp {
String fName;
String lName;
public Emp (String fn, String ln) {
fName = fn;
lName = ln;
}
public String getfName() { return fName; }
public String getlName() { return lName; }
}
and the code fragment:
List<Emp> emp = Arrays.asList (
new Emp ("John", "Smith"),
new Emp ("Peter", "Sam"),
new Emp ("Thomas", "Wale"));
emp.stream()
//line n1
.collect(Collectors.toList());
Which code fragment, when inserted at line n1, sorts the employees list in descending order of fNameand then ascending order of lName?
.sorted (Comparator.comparing(Emp::getfName).reserved().thenComparing
A. .map(Emp::getfName).sorted(Comparator.reserveOrder())
B. (Emp::getlName))
.sorted (Comparator.comparing(Emp::getfName).thenComparing(Emp::getlName))
C. .map(Emp::getfName).sorted(Comparator.reserveOrder().map
D. (Emp::getlName).reserved
Answer: B