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Exam Code: C-THR95-2105

Exam Name: SAP Certified Application Associate - SAP SuccessFactors Career Development Planning and Mentoring 1H/2021

Certification Provider: SAP

Related Certification: SAP Certified Application Associate - SAP SuccessFactors Career Development Planning and Mentoring 1H/2021

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NEW QUESTION: 1
Which of the following would MOST likely ensure that a system development project meets business objectives?
A. User involvement in system specification and acceptance
B. Development of a project plan identifying all development activities
C. Strict deadlines and budgets
D. Development and tests are run by different individuals
Answer: A
Explanation:
Explanation/Reference:
Effective user involvement is the most critical factor in ensuring that the application meets business objectives.
A great way of getting early input from the user community is by using Prototyping. The prototyping method was formally introduced in the early 1980s to combat the perceived weaknesses of the waterfall model with regard to the speed of development. The objective is to build a simplified version (prototype) of the application, release it for review, and use the feedback from the users' review to build a second, better version.
This is repeated until the users are satisfied with the product. t is a four-step process:
initial concept,
design and implement initial prototype,
refine prototype until acceptable, and
complete and release final version.
There is also the Modified Prototype Model (MPM. This is a form of prototyping that is ideal for Web application development. It allows for the basic functionality of a desired system or component to be formally deployed in a quick time frame. The maintenance phase is set to begin after the deployment. The goal is to have the process be flexible enough so the application is not based on the state of the organization at any given time. As the organization grows and the environment changes, the application evolves with it, rather than being frozen in time.
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 12101-12108 and 12099-12101). Auerbach Publications. Kindle Edition.
and
Information Systems Audit and Control Association, Certified Information Systems Auditor 2002 review manual, chapter 6: Business Application System Development, Acquisition, Implementation and Maintenance (page 296).

NEW QUESTION: 2
What is the most important factor in determining sustained bandwidth in an IP SAN with high bandwidth with long delays?
A. Zero-copy TCP
B. TCP sliding window size
C. Ethernet MTU size
D. IP fragmentation
Answer: B

NEW QUESTION: 3
A patient with chronic obstructive pulmonary disease (COPD) is to use a nebulizer. The observation that
indicates the nebulizer is being used incorrectly and additional teaching by the nurse is required would be
that the patient:
A. Inhales with the lips tightly sealed around the mouthpiece of the nebulizer
B. Holds the inspired breath for at least 3 seconds
C. Places the tip of the nebulizer just beyond the lips
D. Exhales slowly through the mouth with the lips pursed slightly
Answer: A
Explanation:
The observation that indicates the nebulizer is being used incorrectly and additional teaching
by the nurse is required would be that the patient inhales with the lips tightly sealed around the
mouthpiece of the nebulizer. This technique results in nasal breathing, which negates the effects of
aerosol medication.

NEW QUESTION: 4
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.255.255.255.254
B. Permit 10.0.0.1.0.0.0.254
C. Permit 10.0.0.1.0.0.0.0
D. Permit 10.0.0.0.0.0.0.1
Answer: B
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 000 ) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.


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