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NEW QUESTION: 1

http://www.supplierdomain.com/status/customerID

A. Amazon SQS; Amazon SNS
B. Amazon S3; Amazon CloudFront
C. Amazon API Gateway; AWS Lambda
D. Amazon ElastiCache; Amazon Elacticsearch Service
E. Elastic Load Balancing; Amazon EC2
Answer: B,C

NEW QUESTION: 2
金融機関のマネーロンダリング防止の専門家は、2004年以降、特定の個人のすべての取引記録を提供するための法的要求を受け取っています。
A. 2004年以降のその個人のすべての電信送金
B. 2004年以降にその個人によって開かれたアカウントの署名カード
C. 2004年以降のその個人のすべての証券取引活動
D. 2004年以降のその個人の月次明細書および取引活動
Answer: A,C,D

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You are the administrator of a computer named Computer1 that runs Windows 7.
Several users share Computer1.
The users plan to encrypt files on the computer.
You need to ensure that you can access all encrypted files on the computer.
What should you do first?
A. At a command prompt, run Certutil.exe.
B. From User Accounts, run the Manage your file encryption certificates wizard.
C. From the local comput er policy, modify the User Rights Assignment.
D. At a command prompt, run Cipher.exe.
Answer: D
Explanation:
Cipher
Displays or alters the encryption of folders and files on NTFS volumes. Used without parameters, cipher displays the encryption state of the current folder and any files it contains.
Administrators can use Cipher.exe to encrypt and decrypt data on drives that use the NTFS file system and to view the encryption status of files and folders from a command prompt. The updated version adds another security option. This new option is the ability to overwrite data that you have deleted so that it cannot be recovered and accessed.
When you delete files or folders, the data is not initially removed from the hard disk. Instead, the space on the disk that was occupied by the deleted data is "deallocated." After it is deallocated, the space is available for use when new data is written to the disk. Until the space is overwritten, it is possible to recover the deleted data by using a low-level disk editor or data-recovery software.
If you create files in plain text and then encrypt them, Encrypting File System (EFS) makes a backup copy of the file so that, if an error occurs during the encryption process, the data is not lost. After the encryption is complete, the backup copy is deleted. As with other deleted files, the data is not completely removed until it has been overwritten. The new version of the Cipher utility is designed to prevent unauthorized recovery of such data.
/K Creates a new certificate and key for use with EFS. If this option is chosen, all the other options will be ignored. By default, /k creates a certificate and key that conform to current group plicy. If ECC is specified, a self-signed certificate will be created with the supplied key size.
/R Generates an EFS recovery key and certificate, then writes them to a .PFX file (containing certificate and private key) and a .CER file (containing only the certificate). An administrator may add the contents of the .CER to the EFS recovery policy to create the recovery for users, and import the .PFX to recover individual files. If SMARTCARD is specified, then writes the recovery key and certificate to a smart card. A .CER file is generated (containing only the certificate). No .PFX file is genereated. By default, /R creates an 2048-bit RSA recovery key and certificate. If EECC is specified, it must be followed by a key size of 356, 384, or 521.

NEW QUESTION: 4
You are a database developer for an application hosted on a Microsoft SQL Server 2012 server.
The database contains two tables that have the following definitions:

Global customers place orders from several countries.
You need to view the country from which each customer has placed the most orders.
Which Transact-SQL query do you use?
A. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY COUNT(OrderAmount) DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
B. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY OrderAmount DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
C. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM
(SELECT c.CustomerID, c.CustomerName, o.ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY COUNT(o.OrderAmount) ASC) AS Rnk
FROM Customer c
INNER JOIN Orders o
ON c.CustomerID = o.CustomerID
GROUP BY c.CustomerID, c.CustomerName, o.ShippingCountry) cs
WHERE Rnk = 1
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
FROM Customer c
INNER JOIN
(SELECT CustomerID, ShippingCountry,
COUNT(OrderAmount) DESC) AS OrderAmount
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
ORDER BY OrderAmount DESC
Answer: A