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1z0-1067-23

Exam Code: 1z0-1067-23

Exam Name: Oracle Cloud Infrastructure 2023 Cloud Operations Professional

Version: V22.75

Q & A: 580 Questions and Answers

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NEW QUESTION: 1
A client wants customers to be able to select a laptop category from a drop-down list in the product configuration
section. After the laptop category is selected, all laptops within that category need to be displayed without the user
having to do any more selections.
Which option meets this requirement?
A. Commerce library function
B. Formula Administration
C. Util library function
D. Auto Update
Answer: D

NEW QUESTION: 2
A customer has a newly installed EMC Avamar Gen4S server with three 3.9 TB nodes in their headquarters data center. They are backing up two database servers with 1 TB of small databases each. The customer requires that the servers can be restored in six hours.
The customer's database administrators need to know that they can meet their RTO. In addition, they want to know how to test the throughput rate when recovering all data and how to format the recovery command to guarantee throughput.
What is a recommendation for testing the throughput rate and recovery?
A. Using a recovery rate of 75 GB/hr/node, two nodes will meet the RTO.Use --validate and
--sequencenumber for throughput testing and recovery.
B. Using a recovery rate of 37.5 GB/hr/node, two nodes will meet the RTO. Use --validate for testing of recovery throughput and --allnodes for recoveries.
C. Using a recovery rate of 37.5 GB/hr/node, three nodes will meet the RTO.Use --validate and
--sequencenumber for throughput testing and recovery.
D. Using a recovery rate of 75 GB/hr/node, three nodes will meet the RTO. Use --validate for testing of recovery throughput and --allnodes for recoveries.
Answer: D

NEW QUESTION: 3
Which of the following statements are true:
I. Heavy tailed parametric distributions are a good choice for severity modeling in operational risk.
II. Heavy tailed body-tail distributions are a good choice for severity modeling in operational risk.
III. Log-likelihood is a means to estimate parameters for a distribution.
IV. Body-tail distributions allow modeling small losses differently from large ones.
A. All of the above
B. II, III and IV
C. I and IV
D. II and III
Answer: A
Explanation:
Explanation
When modeling for operational risk, we are generally concerned with tail losses - this is because the horizon for operational risk is 1 year at the 99.9th percentile. Since the 99.9th percentile is in the tail region, we would like to ensure that the tails are modeled as accurately as possible. Operational risk distributions are modeled using heavy tailed distributions.
Heavy tailed parametric distributions such as log-normal, pareto and others are therefore a good choice for modeling risk severity, therefore statement I is correct.
Body-tail distributions are combinations of parametric distributions, with different types of distributions being used to model the body and the tail - this provides flexibility because small and medium losses upto a threshold can be modeled using one distribution, and losses beyond the threshold can be modeled using a different distribution that is a better estimate of the tail. Statement II is therefore correct.
A log-likelihood function simplifies the optimization of a regular likelihood function. We generally maximize (or minimize the risk functional) a likelihood function with a view to estimating the parameters of the underlying distribution. If the likelihood function is complex, it may sometimes make it mathematically easier to optimize the log of the function - as that changes exponents and multiplications to additions, while behaving in the same way as the underlying function. Therefore statement III is correct, log-likelihood is a means to estimate parameters for a distribution.
Statement IV is correct as body-tail distributions allow modeling different parts of the distribution differently from each other.

NEW QUESTION: 4
View the Exhibit and examine the structure of the CUSTOMERS table.

NEW_CUSTOMERS is a new table with the columns CUST_ID, CUST_NAME and CUST_CITY that have the same data types and size as the corresponding columns in the CUSTOMERS table.
Evaluate the following INSERT statement:

The INSERT statement fails when executed. What could be the reason?
A. The VALUES clause cannot be used in an INSERT with a subquery
B. The WHERE clause cannot be used in a sub query embedded in an INSERT statement
C. Column names in the NEW_CUSTOMERS and CUSTOMERS tables do not match
D. The total number of columns in the NEW_CUSTOMERS table does not match the total number of columns in the CUSTOMERS table
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Copying Rows from Another Table
Write your INSERT statement with a subquery:
Do not use the VALUES clause.
Match the number of columns in the INSERT clause to those in the subquery.
Inserts all the rows returned by the subquery in the table, sales_reps.

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