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NEW QUESTION: 1
An IBM Security QRadar SIEM V7.2.8 Administrator wants to create a security profile within the system but receives an error upon saving.
What is a possible reason for this error?
A. The Administrator has used non alpha numeric value(s) in the name which is not allowed.
B. The Administrator must bring the IBM Security QRadar SIEM V7.2.8 system first in edit mode beforechanges are allowed.
C. The Administrator has mixed non alpha numeric value(s) and alpha numeric value(s) in the name which isnot allowed.
D. The Administrator has used less than 3 characters or more than 30 characters as name of the securityprofile.
Answer: D
Explanation:
In the Security Profile Name field, type a unique name for the security profile. The security profile name mustmeet the following requirements: minimum of 3 characters and maximum of 30 characters.
Reference: ftp://public.dhe.ibm.com/software/security/products/qradar/documents/7.2.1/QRadar/EN/b_qradar_admin_guide.pdf
NEW QUESTION: 2

A. Option B
B. Option A
C. Option D
D. Option C
Answer: A,D
NEW QUESTION: 3
Ted is in a group that has Author access in the Review application's ACL. However, he cannot edit a review he created. Which one of the following is the most likely reason?
A. The developer neglected to include an Authors field to the Review form.
B. He must have Editor or higher access to modify documents.
C. Authors can only create documents.
D. The developer neglected to select the "modify documents" option in the ACL.
Answer: A
NEW QUESTION: 4
What happens when you attempt to compile and run the following code?
# include <vector>
# include <iostream>
# include <algorithm>
using namespace std;
class B { int val;
public:
B(int v):val(v){}
int getV() const {return val;} bool operator < (const B & v) const { return val<v.val;} }; ostream & operator <<(ostream & out, const B & v) { out<<v.getV(); return out;} template<class T>struct Out {
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) { out<<val<<" "; } };
int main() {
B t1[]={3,2,4,1,5};
B t2[]={6,10,8,7,9};
vector<B> v1(10);
sort(t1, t1+5);
sort(t2, t2+5);
merge(t1,t1+5,t2,t2+5,v1.begin());
for_each(v1.begin(), v1.end(), Out<B>(cout));cout<<endl;
return 0;
}
Program outputs:
A. 3 2 4 1 5 6 7 8 9 10
B. 1 2 3 4 5 6 10 8 7 9
C. compilation error
D. 3 2 4 1 5 6 10 8 7 9
E. 1 2 3 4 5 6 7 8 9 10
Answer: C