VALID DUMPS 1Z1-106 FILES & RELIABLE 1Z1-106 SOURCE

Valid Dumps 1z1-106 Files & Reliable 1z1-106 Source

Valid Dumps 1z1-106 Files & Reliable 1z1-106 Source

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Why Getting Certified For Oracle 1Z0-106 Will Be A Smart Move?

The Oracle 1Z0-106 exam is the cert for system administrators who have to work with Oracle databases. This certification is very important if you are a system administrator and want to be an expert in managing Oracle Database server.

The 1Z0-106 exam has been designed to test your knowledge about administration of a database server. It will also test your ability to manage Oracle, SQL, PL/SQL and Java applications as well as UNIX and Linux operating systems.

Oracle 1Z0-106 is a part of the Oracle Certified Professional System Administration program. The certification is valid for five years and it has been divided into two levels: one level for the generalists and another level for the specialists or consultants. The specialist level requires passing all three exams (1Z0-001, 1Z0-002 and 1Z0-003) while the generalist level requires passing only two exams (1Z0-101 and 1Z0-102). Oracle 1Z0-106 exam dumps will help your to pass your exam in your first attempt.

Oracle 1Z0-106 certification exam is intended for professionals who have experience working with Oracle Linux 8 systems. 1z1-106 Exam Tests candidates on various topics, including system administration, networking, security, storage management, and troubleshooting. Candidates should have a solid understanding of Linux command-line tools, file systems, and system services.

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Oracle 1Z0-106 exam is a challenging certification that requires significant preparation and study. Candidates must have a deep understanding of Oracle Linux 8 and be proficient in a range of advanced system administration tasks. 1z1-106 Exam consists of multiple-choice questions and requires candidates to demonstrate their knowledge of the subject matter through practical exercises. Those who pass the exam will have demonstrated their expertise in Oracle Linux 8 Advanced System Administration and will be well-prepared to take on the challenges of managing complex enterprise systems.

Oracle Linux 8 Advanced System Administration Sample Questions (Q19-Q24):

NEW QUESTION # 19
Examine /etc/anacrontab:
SHELL=/bin/sh
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root
# the maximal random delay added to the base delay of the jobs
RANDOM_DELAY=45
# the jobs are started during the following hours only
START_HOURS_RANGE=3-22
# period in days delay in minutes job-identifier command
1 5 dailyjob nice run-parts /etc/cron.daily
7 25 weeklyjob nice run-parts /etc/cron.weekly
@monthly 45 monthlyjob nice run-parts /etc/cron.monthly
Which two statements are true about the jobs scheduled in this file?

  • A. Scripts run by the third job are delayed between 45 and 90 minutes.
  • B. Scripts run by the first job are delayed between 11 and 45 minutes.
  • C. Scripts run by the second job are delayed between 31 and 70 minutes.
  • D. Jobs defined in this anacrontab file can be executed between 15:00 and 22:00
  • E. Jobs defined in this anacrontab file are randomly delayed by up to 51 minutes.

Answer: A,D

Explanation:
Understanding the anacrontab Entries:
The /etc/anacrontab file contains the following variables and job definitions:
SHELL=/bin/sh
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root
# the maximal random delay added to the base delay of the jobs
RANDOM_DELAY=45
# the jobs are started during the following hours only
START_HOURS_RANGE=3-22
# period in days delay in minutes job-identifier command
1 5 dailyjob nice run-parts /etc/cron.daily
7 25 weeklyjob nice run-parts /etc/cron.weekly
@monthly 45 monthlyjob nice run-parts /etc/cron.monthly
Variables Explained:
* RANDOM_DELAY=45
* Specifies that arandom delaybetween 0 and 45 minutes is added to the base delay of each job.
* START_HOURS_RANGE=3-22
* Jobs are allowed to start only between03:00 (3 AM)and22:00 (10 PM).
Job Delays Calculated:
* First Job (dailyjob):
* Base Delay:5 minutes
* Random Delay:0 to 45 minutes
* Total Delay:5 + (0 to 45) =5 to 50 minutes
* Second Job (weeklyjob):
* Base Delay:25 minutes
* Random Delay:0 to 45 minutes
* Total Delay:25 + (0 to 45) =25 to 70 minutes
* Third Job (monthlyjob):
* Base Delay:45 minutes
* Random Delay:0 to 45 minutes
* Total Delay:45 + (0 to 45) =45 to 90 minutes
Option B: Jobs defined in this anacrontab file can be executed between 15:00 and 22:00
* Explanation:
* The START_HOURS_RANGE=3-22 setting allows jobs to start between03:00 and 22:00.
* Therefore, it's true that jobscan be executed between 15:00 (3 PM) and 22:00 (10 PM).
* This statement is correct because the specified time range falls within the allowed start hours.
* Oracle Linux Reference:
* OracleLinux 8: Scheduling Tasks- Section on "Anacron Configuration Files":
"The START_HOURS_RANGE variable defines the time window during which Anacron jobs can run." Option C: Scripts run by the third job are delayed between 45 and 90 minutes.
* Explanation:
* The third job (monthlyjob) has abase delay of 45 minutes.
* With a RANDOM_DELAY of up to 45 minutes, thetotal delaybefore execution is between45 and 90 minutes.
* Therefore, this statement is accurate.
* Oracle Linux Reference:
* OracleLinux 8: Scheduling Tasks- Section on "Understanding Anacron Job Delays":
"Each job's delay is calculated by adding its defined delay to a random value between 0 and RANDOM_DELAY." Why Other Options Are Not Correct:
* Option A:Scripts run by the first job are delayed between 11 and 45 minutes.
* Explanation:
* The first job (dailyjob) has a total delay between5 and 50 minutes(5-minute base delay plus up to 45 minutes random delay).
* Therefore, the statement specifying11 to 45 minutesis incorrect.
* Option D:Jobs defined in this anacrontab file are randomly delayed by up to 51 minutes.
* Explanation:
* The maximum random delay is set by RANDOM_DELAY=45, so the random delay is up to45 minutes, not 51.
* Therefore, this statement is false.
* Option E:Scripts run by the second job are delayed between 31 and 70 minutes.
* Explanation:
* The second job (weeklyjob) has a total delay between25 and 70 minutes(25-minute base delay plus up to 45 minutes random delay).
* The statement specifies a delay between31 and 70 minutes, which is incorrect because the minimum delay is 25 minutes.
Conclusion:
Options B and C are correct based on the configuration specified in /etc/anacrontab. They accurately reflect the possible execution times and delays for the jobs defined.


NEW QUESTION # 20
Examine these commands executed by root:
# mkdir -p /jail /jail/bin /jail/lib64
# cp $(which bash) /jail/bin/
# ldd $(which bash)
linux-vdso.so.1 (0x00007ffd574f5000)
libtinfo.so.6 => /lib64/libtinfo.so.6 (0x00007fb458c2c000)
libdl.so.2 => /lib64/libdl.so.2 (0x00007fb458a28000)
libc.so.6 => /lib64/libc.so.6 (0x00007fb458666000)
/lib64/ld-linux-x86-64.so.2 (0x00007fb459177000)
# cp /lib64/libtinfo.so.6 /jail/lib64/
# cp /lib64/libdl.so.2 /jail/lib64/
# cp /lib64/libc.so.6 /jail/lib64/
# cp /lib64/ld-linux-x86-64.so.2 /jail/lib64/
# chroot /jail
What is the output from the cd, pwd, and ls commands?

  • A. bash-4.4# cd
    bash: cd: /root: No such file or directory
    bash-4.4# pwd
    /root
    bash-4.4# ls
    bash: ls: command not found
  • B. bash-4.4# cd
    bash: cd: /root: Unable to access chrooted file or directory /root
    bash-4.4# pwd
    /
    bash-4.4# ls
    bin lib64
  • C. bash-4.4# cd
    bash: cd: /root: No such file or directory
    bash-4.4# pwd
    /
    bash-4.4# ls
    bin lib64
  • D. bash-4.4# cd
    bash: cd: command not found
    bash-4.4# pwd
    bash: pwd: command not found
    bash-4.4# ls
    bash: ls: command not found

Answer: C


NEW QUESTION # 21
Which two default user account settings are contained in /etc/login.defs?

  • A. Group hashed passwords.
  • B. User hashed passwords.
  • C. Decryption method used to decrypt passwords.
  • D. Password aging controls.
  • E. Encryption method used to encrypt passwords.

Answer: D,E

Explanation:
Explanation of Answer D:The/etc/login.defsfile in Oracle Linux contains configuration settings related to user account policies, including password aging controls. This includes settings such asPASS_MAX_DAYS, PASS_MIN_DAYS, andPASS_WARN_AGE, which define the maximum number of days a password is valid, the minimum number of days between password changes, and the number of days before password expiration to warn users, respectively.
Explanation of Answer E:The/etc/login.defsfile also contains settings for the encryption method used to encrypt user passwords. TheENCRYPT_METHODparameter specifies the hashing algorithm, such as SHA512, that is used to encrypt user passwords stored in/etc/shadow.


NEW QUESTION # 22
Which two statements are true about removing a physical volume (PV) from a volume group (VG)?

  • A. It can be removed when an active VG has mounted file systems by running vgexport.
  • B. It can be removed when it is part of an active VG.
  • C. It cannot be removed when it is part of an active VG.
  • D. It can be removed when an inactive logical volume is on the VG.
  • E. It can be removed only after removing it from its VG by using vgreduce.

Answer: C,E


NEW QUESTION # 23
Which two components are used for creating a new rsyslog rule?

  • A. module
  • B. parser
  • C. security policy
  • D. filter
  • E. action

Answer: D,E


NEW QUESTION # 24
......

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