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NEW QUESTION: 1
In the context of access control, locks, gates, guards are examples of which of the following?
A. Administrative controls
B. Technical controls
C. Physical controls
D. Logical controls
Answer: C
Explanation:
Administrative, technical and physical controls are categories of access control mechanisms.
Logical and Technical controls are synonymous. So both of them could be eliminated as possible choices.
Physical Controls: These are controls to protect the organization's people and physical environment, such as locks, gates, and guards. Physical controls may be called "operational controls" in some contexts.
Physical security covers a broad spectrum of controls to protect the physical assets (primarily the people) in an organization. Physical Controls are sometimes referred to as "operational" controls in some risk management frameworks. These controls range from doors, locks, and windows to environment controls, construction standards, and guards. Typically, physical security is based on the notion of establishing security zones or concentric areas within a facility that require increased security as you get closer to the valuable assets inside the facility. Security zones are the physical representation of the defense-in-depth principle discussed earlier in this chapter. Typically, security zones are associated with rooms, offices, floors, or smaller elements, such as a cabinet or storage locker. The design of the physical security controls within the facility must take into account the protection of the asset as well as the individuals working in that area.
Reference(s) used for this question:
Hernandez CISSP, Steven (2012-12-21). Official (ISC)2 Guide to the CISSP CBK, Third Edition ((ISC)2 Press) (Kindle Locations 1301-1303). Auerbach Publications. Kindle Edition. and Hernandez CISSP, Steven (2012-12-21). Official (ISC)2 Guide to the CISSP CBK, Third Edition ((ISC)2 Press) (Kindle Locations 1312-1318). Auerbach Publications. Kindle Edition.
NEW QUESTION: 2
What is the first step suggested before implementing any single CIS Control?
A. Perform a vulnerability scan
B. Develop a roll-out schedule
C. Perform a gap analysis
D. Develop an effectiveness test
Answer: C
NEW QUESTION: 3
Which statement is true regarding inline data deduplication and client-side data deduplication?
A. Inline server data deduplication is adapted from the ProtecTier data deduplication algorithm and is dierent from the client side deduplication algorithm
B. Client side deduplication will influence inline server data deduplication and chunks will be optimized
C. Client side deduplication has no influence on inline server deduplication, because the data will be reorganized in container pool structure
D. inline server data deduplication uses the same data deduplicationxthat is used byclient- side data deduplication and existing server-side data deduplication with the IDENTIFY process.
Answer: D
Explanation:
Does the inline data deduplication use the same data deduplication methods as client-side data deduplication? Yes, inline server data deduplication uses the same data deduplication algorithms that is used by client-side data deduplication and existing server-side data deduplication with the IDENTIFY process.
References: https://www.ibm.com/developerworks/community/wikis/home?lang=en#!/wiki/Tivoli%20Storage%20Manager/page/Directory-container%20storage%20pools%20FAQs
NEW QUESTION: 4
The SAS data set ONE contains fifty million observations and contains the variable
PRICE, QUANTITY, FIXED and VARIABLE. Which SAS program successfully creates three new variables TOTREV, TOTCOST and PROFIT and requires the least amount of CPU resources to be processed?
A. data two;
Set one;
Where totrev>1000;
Totrev=sum(price*quantity);
Totcost=sum(fixed,variable);
Profit=sum(totrev,-totcost);
Run;
B. data two;
Set one;
Totrev=sum(price*quantity);
If totrev>1000;
Totcost=sum(fixed,variable);
Profit=sum(totrev,-totcost);
Run;
C. data two;
Set one;
Totrev = sum(price*quantity);
Totcost= sum(fixed,variable);
If totrev>1000;
Profit=sum(totrev,-totcost);
Run;
D. data two;
Set one;
totrev=sum(price*quantity);
where totrev>1000;
totcost=sum(fixed,variable);
profit=sum(totrev,-totcost);
run;
Answer: B