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NEW QUESTION: 1
CORRECT TEXT
User A belongs to profile 'Custom - Recruiter'. At the profile level (Path as per Enhanced Profile User Interface, Setup -> Manage Users -> Profiles -> Custom - Recruiter -> Objects and Tabs -> Accounts) he does not have Read access on Account object. But yet he is able to view account records. He is not able to edit or delete account records. What could be the reason?
Answer:
Explanation:
The
profile 'Custom - Recruiter' has 'View All Data' enabled at this path,Setup -> Manage Users -> Profiles -> Custom - Recruiter -> System Permissions. (Path is as per Enhanced Profile User Interface). This gives all users belonging to this profile the ability to view any type of data across all apps.
NEW QUESTION: 2
멀티미디어 회사는 가입자에게 VOD (Video-on-demand) 콘텐츠를 비용 효율적인 방식으로 제공해야 합니다. 비디오 파일의 크기는 1-15GB이며 일반적으로 생성 후 처음 6 개월 동안 자주 본 다음 액세스가 상당히 감소합니다. 회사는 구독자가 즉시 사용할 수 있도록 모든 비디오 파일을 보답합니다. 현재 약 30.000 개의 파일이 있으며 회사는 시간이 지남에 따라 이 수를 두 배로 늘릴 것으로 예상합니다.
회사의 VOD 콘텐츠를 제공하기 위한 가장 비용 효율적인 솔루션은 무엇입니까?
A. Amazon S3 Standard에 비디오 파일을 저장합니다. 비디오 파일을 6 개월 후 S3 Standard-Infrequent Access (S3 Standard-IA)로 이동하고 1 년 후 S3 Glacier Deep Archive로 이동하는 S3 수명주기 규칙을 생성합니다. Amazon CloudFront를 사용하여 S3 버킷을 오리진으로 사용하여 콘텐츠를 전송합니다.
B. AWS Elemental MediaConvert를 사용하고 적응 형 비트 전송률 비디오 파일을 Amazon S3에 저장합니다. Amazon S3에서 콘텐츠를 전송하도록 AWS Elemental MediaPackage 엔드 포인트를 구성합니다.
C. Amazon Elastic File System (Amazon EFS) Standard에 비디오 파일을 저장합니다. EFS 수명주기 관리를 활성화하여 6 개월 후에 비디오 파일을 EFS Infrequent Access로 이동합니다. Elastic Load Balancer 뒤에 Amazon EC2 Auto Scaling 그룹을 생성하여 Amazon EFS에서 콘텐츠를 전송합니다.
D. S3 Intelligent-Tiering을 사용하여 Amazon S3 버킷에 비디오 파일을 저장합니다. Amazon CloudFront를 사용하여 S3 버킷을 오리진으로 사용하여 콘텐츠를 전송합니다.
Answer: D
NEW QUESTION: 3
View the Exhibit and examine the structure of the CUSTOMERS table.
You want to generate a report showing the last names and credit limits of all customers
whose last names start with A, B, or C, and credit limit is below 10, 000.
Evaluate the following two queries:
Which statement is true regarding the execution of the above queries?
A. Only the first query gives the correct result.
B. Only the second query gives the correct result.
C. Both execute successfully but do not give the required result.
D. Both execute successfully and give the same result.
Answer: A
NEW QUESTION: 4
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.0.0.0.1
B. 許可10.0.0.1.0.0.0.0
C. 許可10.0.0.0.255.255.255.254
D. 許可10.0.0.1.0.0.0.254
Answer: D
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's 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 ranges from 172.23.16.0 to 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 0000) 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.