Juniper Latest JN0-224 Practice Materials, Latest Braindumps JN0-224 Ppt | JN0-224 Certification Questions - Cads-Group

  • Exam Number/Code : JN0-224
  • Exam Name : Automation and DevOps, Associate (JNCIA-DevOps)
  • Questions and Answers : 260 Q&As
  • Price: $ 99.00 $ 39.00

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NEW QUESTION: 1
Which of the following is NOT an administrative control?
A. Change control procedures
B. Development of policies, standards, procedures and guidelines
C. Logical access control mechanisms
D. Screening of personnel
Answer: C
Explanation:
It is considered to be a technical control.
Logical is synonymous with Technical Control. That was the easy answer.
There are three broad categories of access control: Administrative, Technical, and Physical.
Each category has different access control mechanisms that can be carried out manually or automatically. All of these access control mechanisms should work in concert with each other to protect an infrastructure and its data.
Each category of access control has several components that fall within it, as shown here:
Administrative Controls
Policy and procedures
Personnel controls
Supervisory structure
Security-awareness training
Testing
Physical Controls
Network segregation Perimeter security Computer controls Work area separation Data backups
Technical Controls
System access Network architecture Network access Encryption and protocols Control zone Auditing
The following answers are incorrect :
Screening of personnel is considered to be an administrative control
Development of policies, standards, procedures and guidelines is considered to be an administrative control
Change control procedures is considered to be an administrative control. Reference : Shon Harris AIO v3 , Chapter - 3 : Security Management Practices , Page : 5254

NEW QUESTION: 2
ベンダーのセキュリティ要件を満たす必要があります。
あなたは何をするべきか?
A. Azure Cloud ShellからSet-MsolUserPrincipalNameを実行し、-tenantIDパラメーターを指定します。
B. Azure Cloud Shellから、New-AzureADMSInvitationを実行し、-InvitedIserEmailAddressコマンドレットを指定します。
C. Azureポータルから、IDプロバイダーを追加します。
D. Azureポータルから、認証方法を変更します。
Answer: B
Explanation:
Explanation
* Vendors must be able to authenticate by using their Microsoft account when accessing Contoso
* resources.
You can invite guest users to the directory, to a group, or to an application. After you invite a user through any of these methods, the invited user's account is added to Azure Active Directory (Azure AD), with a user type of Guest. The guest user must then redeem their invitation to access resources. An invitation of a user does not expire.
The invitation will include a link to create a Microsoft account. The user can then authenticate using their Microsoft account. In this question, the vendors already have Microsoft accounts so they can authenticate using them.
In this solution, we are creating guest account invitations by using the New-AzureADMSInvitation cmdlet and specifying the -InvitedUserEmailAddress parameter.
Reference:
https://docs.microsoft.com/en-us/azure/active-directory/b2b/add-users-administrator
https://docs.microsoft.com/en-us/powershell/module/azuread/new-azureadmsinvitation?view=azureadps-2.0

NEW QUESTION: 3
teeコマンドのどのオプションを使用すると、既存のファイルの内容を上書きする代わりに、出力ファイルの最後に出力が連結されますか?
A. --no-clobber
B. -a
C. --continue
D. -c
Answer: B

NEW QUESTION: 4
The following excerpt is taken from a honeyput log. The log captures activities across three days. There are several intrusion attempts; however, a few are successful. Study the log given below and answer the following question:
(Note: The objective of this questions is to test whether the student has learnt about passive OS fingerprinting (which should tell them the OS from log captures): can they tell a SQL injection attack signature; can they infer if a user ID has been created by an attacker and whether they can read plain source - destination entries from log entries.)

What can you infer from the above log?
A. The system is a web application server compromised through SQL injection.
B. The system has been compromised and backdoored by the attacker.
C. The system is a windows system which is being scanned unsuccessfully.
D. The actual IP of the successful attacker is 24.9.255.53.
Answer: C

 

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