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NEW QUESTION: 1
A database supports three applications: CRM, ERP, and ACC. These applications connect to the database by using three different services: CRM_SRV for the CRM application, ERP_SRV for the ERP application, and ACC_SRV for the ACC application.
You enable tracing for the ACC_SRV service by issuing the following command:
SQL> EXECUTE DBMS for the ACC_SRV service by issuing the following command:
SQL> EXECUITIVE DBMS_MONITOR. SERV_MOD_ACT_TRACE_ENABLE (service_name => 'ACC_SRV', waits => TRUE, binds = > FALSE, instance_name = > 'inst1');
Which statement is true?
A. A single trace file is created for each session that uses the ACC_SRV service.
B. All trace information for the service connection to inst1 will be stored in a single trace file.
C. All trace information for the ACC_SRV service connected to inst1 is stored in multiple trace files, which can be consolidated by using the tkprof utility.
D. A trace file is not created because the module name is not specified.
E. Only those SQL statements that are identified with the ACC_SRV service executed on the inst1 instance are recorded in trace files.
Answer: A
Explanation:
SERV_MOD_ACT_TRACE_ENABLE
serv_mod_act_trace_enable and serv_mod_act_trace_disable, which enables and disables
trace for given service_name, module and action.
For example for a given service name you can trace all session started from SQL*Plus.
Module and action in your own created application can be set using dbms_application_info
set_module and set_action procedures.
serv_mod_act_trace_enable fills sys table wri$_tracing_enabled and view
dba_enabled_traces on top of this table as follows:
SQL> exec dbms_monitor.serv_mod_act_trace_enable(service_name=>'orcl',
module_name=>'SQL*Plus')
PL/SQL procedure successfully completed.
SQL> select * from sys.wri$_tracing_enabled;
TRACE_TYPE PRIMARY_ID QUALIFIER_ID1 QUALIFIER_ID2 INSTANCE_NAME
FLAGS
4 orcl SQL*Plus 8 SQL> select * from dba_enabled_traces;
TRACE_TYPE PRIMARY_ID QUALIFIER_ID1 QUALIFIER_ID2 WAITS BINDS INSTANCE_NAME
SERVICE_MODULE orcl SQL*Plus TRUE FALSE

NEW QUESTION: 2
JS Industries has expanded their business with the addition of their first remote office. The remote office router (R3) was previously configured and all corporate subnets were reachable from R3. JS Industries is interested in using route summarization along with the EIGRP Stub Routing feature to increase network stability while reducing the memory usage and bandwidth utilization to R3. Another network professional was tasked with implementing this solution. However, in the process of configuring EIGRP stub routing connectivity with the remote network devices off of R3 has been lost.
Currently EIGRP is configured on all routers R2, R3, and R4 in the network. Your task is to identify and resolve the cause of connectivity failure with the remote office router R3. Once the issue has been resolved you should complete the task by configuring route summarization only to the remote office router R3.
You have corrected the fault when pings from R2 to the R3 LAN interface are successful, and the R3 IP routing table only contains 2 10.0.0.0 subnets.


Answer:
Explanation:
Here are the solution as below:
Explanation:
First we have to figure out why R3 and R4 can not communicate with each other. Use the show running- config command on router R3.

Notice that R3 is configured as a stub receive-only router. The receive-only keyword will restrict the router from sharing any of its routes with any other router in that EIGRP autonomous system. This keyword will also prevent any type of route from being sent. Therefore we will remove this command and replace it with the eigrp stub command:
R3# configure terminal R3(config)# router eigrp 123 R3(config-router)# no eigrp stub receive-only R3(config-router)# eigrp stub R3(config-router)# end Now R3 will send updates containing its connected and summary routes to other routers. Notice that the eigrp stub command equals to the eigrp stub connected summary because the connected and summary options are enabled by default.
Next we will configure router R3 so that it has only 2 subnets of 10.0.0.0 network. Use the show ip route command on R3 to view its routing table:

Because we want the routing table of R3 only have 2 subnets so we have to summary sub-networks at the interface which is connected with R3, the s0/0 interface of R4.
There is one interesting thing about the output of the show ip route shown above: the 10.2.3.0/24, which is a directly connected network of R3. We can't get rid of it in the routing table no matter what technique we use to summary the networks. Therefore, to make the routing table of R3 has only 2 subnets we have to summary other subnets into one subnet.
In the output if we don't see the summary line (like 10.0.0.0/8 is a summary...) then we should use the command ip summary-address eigrp 123 10.2.0.0 255.255.0.0 so that all the ping can work well.
In conclusion, we will use the ip summary-address eigrp 123 10.2.0.0 255.255.0.0 at the interface s0/0 of R4 to summary.
R4> enable R4# conf t
R4(config)# interface s0/0 R4(config-if)# ip summary-address eigrp 123 10.2.0.0 255.255.0.0 Now we jump back to R3 and use the show ip route command to verify the effect, the output is shown below:

Note: Please notice that the IP addresses and the subnet masks in your real exam might be different so you might use different ones to solve this question.
Just for your information, notice that if you use another network than 10.0.0.0/8 to summary, for example, if you use the command ip summary-address eigrp 123 10.2.0.0 255.255.0.0 you will leave a
/16 network in the output of the show ip route command.

But in your real exam, if you don't see the line "10.0.0.0/8 is a summary, Null0" then you can summarize using the network 10.2.0.0/16. This summarization is better because all the pings can work well.
Finally don't forget to use the copy run start command on routers R3 and R4 to save the configurations.
R3(config-if)# end
R3# copy run start
R4(config-if)# end
R4# copy run start
If the "copy run start" command doesn't work then use "write memory."

NEW QUESTION: 3
College Honor Society는 홈 풋볼 게임에서 뜨거운 프레즐을 판매합니다. 프레즐은 각각 $ 1.00에 판매되며 프레첼 당 비용은 $ .30입니다. 판매되지 않은 프레즐은 다음 홈 경기 이전에 유효하지 않기 때문에 버려집니다.
게임당 프레즐 수요의 빈도 분포는 다음과 같습니다.
판매량 확률 2,000 프레즐 .10 3,000 프레즐 .15 4,000 프레즐 .20
5.000
프레즐 .35
6.000
프레즐 .20
가장 가능성있는 결과를 기반으로 결정 론적 접근 방식을 사용하는 다음 홈 풋볼 게임에서 프레즐에 대한 예상 수요는 다음과 같습니다.
A. 4,000 pretzels.
B. 5.000 pretzels.
C. 6,000 pretzels.
D. 4,400 pretzels.
Answer: B
Explanation:
A deterministic approach assumes that a value is known with certainty. If that value is deemed to be the most likely outcome, assumed demand will be 5,000 pretzels, the volume with the highest probability (35%).

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