Exam C_THR82_2311 Review, SAP C_THR82_2311 Valid Exam Fee | C_THR82_2311 Real Brain Dumps - Sapsam

  • Exam Code: C_THR82_2311
  • Exam Name: SAP Certified Application Associate - SAP SuccessFactors Performance and Goals 2H/2023
  • Version: V15.35     Q & A: 208 Questions and Answers

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NEW QUESTION: 1
In Microsoft Dynamics AX 2012, what can you classify by using the direct outsourcing cost group?
A. manufacturing operational resources
B. direct materials
C. surcharges and rates
D. service-type products
Answer: D

NEW QUESTION: 2
Which of the following steps will you take to add a prerecorded audio comment in a PDF?
A. Choose Comment > Drawing Markups > Record Audio.
B. Choose Tools > Annotations > Record Audio.
C. Choose Tools > Drawing Markups > Record Audio.
D. Choose Comment > Annotations > Record Audio.
Answer: D
Explanation:
Take the following steps to add a prerecorded audio comment in a PDF.
1.Choose Comment > Annotations > Record Audio, and then click in the PDF where you want to place the audio comment.

2.Click Browse (Windows) or Choose (Mac OS) for selecting the audio file to be added.
3.(Optional) In order to hear the audio comment, click the Play button .

4.Specify options in the Properties dialog box, and then click the OK button.

NEW QUESTION: 3
-- Exhibit -[edit] user@srx# run show route
inet.0: 10 destinations, 10 routes (10 active, 0 holddown, 0 hidden)
+ = Active Route, - = Last Active, * = Both
0.0.0.0/0 *[Static/5] 01:09:08 > to 172.18.1.1 via ge-0/0/3.0 10.210.14.128/27 *[Direct/0] 8w6d 15:43:09 > via ge-0/0/0.0 10.210.14.135/32 *[Local/0] 11w0d 06:43:04 Local via ge-0/0/0.0 172.18.1.0/30 *[Direct/0] 8w6d 15:43:01 > via ge-0/0/3.0 172.18.1.2/32 *[Local/0] 11w0d 06:43:03 Local via ge-0/0/3.0 172.19.1.0/24 *[Direct/0] 03:46:56 > via ge-0/0/1.0 172.19.1.1/32 *[Local/0] 03:46:56 Local via ge-0/0/1.0 172.20.105.0/24 *[Direct/0] 03:46:56 > via ge-0/0/4.105 172.20.105.1/32 *[Local/0] 03:46:56 Local via ge-0/0/4.105 192.168.30.1/32 *[Direct/0] 4d 03:44:41 > via lo0.0
fbf.inet.0: 2 destinations, 2 routes (2 active, 0 holddown, 0 hidden)
+ = Active Route, - = Last Active, * = Both 0.0.0.0/0 *[Static/5] 00:00:11 > to 172.19.1.2 via ge-0/0/1.0 172.19.1.0/24 *[Direct/0] 00:00:11 > via ge-0/0/1.0
[edit]
user@srx# show routing-instances
fbf {
routing-options {
static {
route 0.0.0.0/0 next-hop 172.19.1.2;
}
}
}
[edit]
user@srx# show routing-options
interface-routes {
rib-group inet fbf-int;
}
static {
route 0.0.0.0/0 next-hop 172.18.1.1;
}
rib-groups {
fbf-int {
import-rib [ inet.0 fbf.inet.0 ];
import-policy fbf-pol;
}
}
[edit]
user@srx# show policy-options policy-statement fbf-pol
term 1 {
from interface ge-0/0/1.0;
to rib fbf.inet.0;
then accept;
}
term 2 {
then reject;
}
-- Exhibit -
Referring to the exhibit, you notice that filter-based forwarding is not working.
What is the reason for this behavior?
A. The RIB group is configured incorrectly.
B. The routing instance is configured incorrectly.
C. The default static routes are configured incorrectly.
D. The routing policy is configured incorrectly.
Answer: B
Explanation:
Bydefault, wehave a static route in a routing instancesendingthe default route to 172.19.1.2.Wewant to hijack traffic matching a particular filter and send the traffic to a different next-hop, 172.18.1.1. Weshouldcreate your rib group by importing FIRST the table belonging to your virtual router and SECOND the table for the forwarding instancethat has the next-hop specified.
Reference:http://kb.juniper.net/InfoCenter/index?page=content&id=KB17223

NEW QUESTION: 4
What happens when you attempt to compile and run the following code?
#include <vector>
#include <set>
#include <iostream>
#include <algorithm>
using namespace std;
template<class T>struct Out {
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) { out<<val<<" "; } };
struct Sequence {
int start;
Sequence(int start):start(start){}
int operator()() { return start++; } };
int main() {
vector<int> v1(10);
generate_n(v1.begin(), 10, Sequence(1));
random_shuffle(v1.rbegin(), v1.rend());
sort(v1.begin(), v1.end(), great<int>());
for_each(v1.begin(), v1.end(), Out<int>(cout));cout<<endl;
return 0;
}
Program outputs:
A. 1 2 3 4 5 6 7 8 9 10
B. 8 10 5 1 4 6 2 7 9 3
C. compilation error
D. 10 9 8 7 6 5 4 3 2 1
Answer: C

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