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Vendor: Linux Foundation

Exam Code: PCA Dumps

Questions and Answers: 104

Product Price: $69.00

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NEW QUESTION: 1
In which of the following system development life cycle (SDLC) phases are access control and encryption algorithms chosen?
A. System design specifications
B. Software development
C. Architectural design
D. Procedural design
Answer: A
Explanation:
Explanation
The system design specifications phase is when security specifications are identified. The procedural design converts structural components into a procedural description of the software. The architectural design is the phase that identifies the overall system design, hut not the specifics. Software development is too late a stage since this is the phase when the system is already being coded.

NEW QUESTION: 2
Aktive RFID-Tags (Radio Frequency ID) sind welchen der folgenden Expositionen ausgesetzt?
A. Abhören
B. Schädlicher Code
C. Phishing
D. Sitzungsentführung
Answer: A
Explanation:
Erläuterung:
Aktive RFID-Tags unterliegen wie drahtlose Geräte dem Abhören. Sie unterliegen von Natur aus keinem Session-Hijacking, Schadcode oder Phishing.

NEW QUESTION: 3
You are configuring serverless computing in Azure.
You need to receive an email message whenever a resource is created in or deleted from a resource group.
Which three actions should you perform in sequence? To answer, move the appropriate actions from the list of actions to the answer area and arrange them in the correct order.

Answer:
Explanation:

Explanation

References:
https://docs.microsoft.com/en-us/azure/event-grid/monitor-virtual-machine-changes-event-grid-logic-app

NEW QUESTION: 4
Consider the Information Engineering diagram shown in the exhibit. Building_ID, R_ID, Room_Count and Room_Num are integer numbers, whereas Bldg_Name, Location and Res_Name are all represented by variable-length strings with a maximum of 20 characters.
Which SQL statement best implements the RESIDENT relation shown in this diagram?

A. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumINTEGER,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL);
B. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumINTEGER,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
C. CREATE TABLE RESIDENT (
R_ID INTEGER NOT NULL PRIMARY KEY,
Room_NumBINARY,
Res_NameVARCHAR (20),
Building_IDINTEGER NOT NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
D. CREATE TABLE RESIDENT (
R_ID INTEGER NULL PRIMARY KEY,
Room_NumFLOAT,
Res_NameVARCHAR,
Building_IDINTEGER NULL,
FOREIGN KEY Building_ID REFERENCES BUILDING (Building_ID));
Answer: B


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