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Exam Code: CPHIMS
Exam Name: HIMSS Certified Professional in Healthcare Information and Management Systems
Version: V22.75
Q & A: 580 Questions and Answers
CPHIMS Free Demo download
About HIMSS CPHIMS Exam
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NEW QUESTION: 1
Refer to the exhibit.
The exhibit shows the current plan for a wired network upgrade.
As much as possible, the customer wants to flatten the architecture and avoid recabling. However, each Building 2 switch must also maintain connectivity to the core if one link fails. What should the architect propose to meet the customer requirements?
A. Combine the switches in each Building 2 closet as a VSF fabric; establish two 10 GbE connections to the core per fabric.
B. Use two additional 2930F switches to act as an aggregation layer for Building 2; connect them to the core on 40 GbE connections.
C. Extend additional fiber between the buildings so that each Building 2 switch can have a direct 10 GbE connection to the core.
D. Connect each Building 2 switch directly to the core on a single fiber strand through the use of SFP+-SR transceivers.
Answer: A
NEW QUESTION: 2
CREATE MASTER KEY ENCRYPTION BY PASSWORD = 'MyPassword1!'
CREATE CERTIFICATE TDE_Certificate WITH SUBJECT = 'TDE Certificate';
BACKUP CERTIFICATE TDE_Certificate TO FILE = ''d:\TDE_Certificate.cer'
WITH PRIVATE KEY (FILE = 'D:\TDE_Certificate.key', ENCRYPTION BY PASSWORD
'MyPassword1!');
CREATE DATABASE ENCRYPTION KEY
WITH ALGORITHM = AES_256
ENCRYPTION BY SERVER CERTIFICATE TDE_Certificate;
ALTER DATABASE Orders SET ENCRYPTION ON;
A. CREATE CERTIFICATE TDE_CertificateWITH SUBJECT = 'TDE Certificate';USE Orders;CREATE DATABASE ENCRYPTION KEYWITH ALGORITHM = AES_256ENCRYPTION BY SERVER CERTIFICATE TDE_Certificate;
B. CREATE CERTIFICATE TDE_CertificateFROM FILE = 'd:\TDE_Certificate.cer';
C. ALTER DATABASE MasterSET ENCRYPTION OFF;
D. CREATE CERTIFICATE TDE_CertificateFROM FILE = 'd:\TDE_Certificate.cer'WITH PRIVATE KEY (FILE = 'D:\TDE_Certificate.key',DECRYPTION BY PASSWORD
'MyPassword1!');
Answer: D
NEW QUESTION: 3
展示を表示して、EMPLOYEESテーブルの構造を調べます。
MANAGER_IDが管理する100人の従業員とそのマネージャーを選択するとします。
2つの列に出力が必要です。最初の列には従業員のマネージャーのLAST_NAMEが含まれ、2番目の列には従業員のLAST_NAMEが含まれる必要があります。
どのSQLステートメントを実行しますか?
A. SELECT m.last_name "Manager"、e.last_name "Employee" FROM従業員m JOIN従業員eON e.employee_id = m.manager_idWHERE m.manager_id = 100;
B. SELECT m.last_name "Manager"、e.last_name "Employee" FROM従業員m JOIN従業員eON m.employee_id = e.manager_idWHERE e.manager_id = 100;
C. SELECT m.last_name "Manager"、e.last_name "Employee" FROM従業員m JOIN従業員eON m.employee_id = e.manager_idWHERE m.manager_id = 100;
D. SELECT m.last_name "Manager"、e.last_name "Employee" FROM従業員m JOIN従業員eWHERE m.employee_id = e.manager_id AND e.manager_id = 100
Answer: B
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