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712-50 -- Designing Blue Prism Process Solutions

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NEW QUESTION: 1
Given the following requirements:
Create a table to contain employee data, with a unique numeric identifier automatically assigned when a row is added, has an EDLEVEL column that permits only the values 'C', 'H' and 'N', and permits inserts only when a corresponding value for the employee's department exists in the DEPARTMENT table.
Which of the following CREATE statements will successfully create this table?
A. CREATE TABLEemp (
empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3),
edlevel CHAR(1),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department
(deptno),
CONSTRAINT edlevel_ck CHECK edlevel VALUES ('C','H','N')
);
B. CREATE TABLEemp (
empno SMALLINT NEXTVAL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3) NOT NULL,
edlevel CHAR(1) CHECK IN ('C','H','N')),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY department (deptno) REFERENCES
(workdept)
);
C. CREATE TABLEemp (
empno SMALLINT NEXTVAL GENERATED ALWAYS AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3) NOT NULL,
edlevel CHAR(1),
PRIMARY KEY emp_pk (empno),
FOREIGN KEY emp_workdept_fk ON (workdept) REFERENCES department (deptno),
CHECK edlevel_ck VALUES (edlevel IN ('C','H','N')),
);
D. CREATE TABLEemp (
empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3),
edlevel CHAR(1),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department
(deptno),
CONSTRAINT edlevel_ck CHECK (edlevel IN ('C','H','N'))
);
Answer: D

NEW QUESTION: 2

A. Option B
B. Option C
C. Option A
D. Option D
Answer: D
Explanation:
Explanation
http://www.cisco.com/en/US/technologies/tk648/tk872/technologies_white_paper0900aecd8026003d.pdf One of the key advantages IPv6 brings is the exponentially larger address space. The following will outline the basic address architecture of IPv6.
128-bit-long addresses
Represented in hexadecimal format:
Uses CIDR principles: prefix/prefix length
x:x:x:x:x:x:x:x, where x is a 16-bit hex field
The last 64 bits are used for the interface ID

NEW QUESTION: 3
Which three ways can membership be defined in a dynamic security group? (Choose three.)
A. Security Tags
B. Locale ID
C. Distributed Firewall Rules
D. Regular Expressions
E. Security Groups
Answer: A,B,E
Explanation:
Explanation/Reference:
Reference: https://pubs.vmware.com/NSX-6/index.jsp?topic=%2Fcom.vmware.nsx.admin.doc%2FGUID- B9FC0D05-BE96-4D83-8C58-98B0F96DB342.html

NEW QUESTION: 4
Kyle is performing the final testing of an application he developed for the accounting department. His last round of testing is to ensure that the program is as secure as possible.
Kyle runs the following command. What is he testing at this point? #include #include int main(int argc, char *argv[]) { char buffer[10]; if (argc < 2) { fprintf (stderr, "USAGE: %s string\n", argv[0]); return 1; } strcpy(buffer, argv[1]); return 0; }
A. Buffer overflow
B. Format string bug
C. Kernal injection
D. SQL injection
Answer: A