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NEW QUESTION: 1
Max number of dataflow definitions (with data sync enabled)
A. 0
B. 1
C. 2
D. 3
Answer: D

NEW QUESTION: 2
インプリメンテーショングループは、IPv6のためにconcept1証拠をするために、テストベッドを使っています。計画に対処していて、送っているネットワークのいくつかの変更の後、チケットがそうであったトラブルはそれを示すことを開けました。そして、ループバックはR1の上で申し出ます（2026：111:1）DSW2（2026：102:1）の上でループバック・アドレスの疎通を確認することができません。
障害状態がどの技術に関連していますか？
A. NTP
B. IPV4 and IPV6 Interoperability
C. IPv4 layer 3 security
D. IPv6 OSPF Routing
E. IPv4 OSPF Routing
Answer: B
Explanation:
As explained earlier, the problem is with route misconfigured tunnel modes on R3. R3 is using tunnel mode ipv6, while R4 is using the default of GRE.
Topic 9, Ticket 1: Switch Port Trunk
Topology Overview
(Actual Troubleshooting lab design is for below network design)
Client Should have IP 10.2.1.3
EIGRP 100 is running between switch DSW1 & DSW2
OSPF (Process ID 1) is running between R1, R2, R3, R4
Network of OSPF is redistributed in EIGRP
BGP 65001 is configured on R1 with Webserver cloud AS 65002
HSRP is running between DSW1 & DSW2 Switches
The company has created the test bed shown in the layer 2 and layer 3 topology exhibits.
This network consists of four routers, two layer 3 switches and two layer 2 switches.
In the IPv4 layer 3 topology, R1, R2, R3, and R4 are running OSPF with an OSPF process number 1.
DSW1, DSW2 and R4 are running EIGRP with an AS of 10. Redistribution is enabled where necessary.
R1 is running a BGP AS with a number of 65001. This AS has an eBGP connection to AS 65002 in the ISP's network. Because the company's address space is in the private range.
R1 is also providing NAT translations between the inside (10.1.0.0/16 & 10.2.0.0/16) networks and outside (209.65.0.0/24) network.
ASW1 and ASW2 are layer 2 switches.
NTP is enabled on all devices with 209.65.200.226 serving as the master clock source.
The client workstations receive their IP address and default gateway via R4's DHCP server.
The default gateway address of 10.2.1.254 is the IP address of HSRP group 10 which is running on DSW1 and DSW2.
In the IPv6 layer 3 topology R1, R2, and R3 are running OSPFv3 with an OSPF process number 6.
DSW1, DSW2 and R4 are running RIPng process name RIP_ZONE.
The two IPv6 routing domains, OSPF 6 and RIPng are connected via GRE tunnel running over the underlying IPv4 OSPF domain. Redistrution is enabled where necessary.
Recently the implementation group has been using the test bed to do a 'proof-of-concept' on several implementations. This involved changing the configuration on one or more of the devices. You will be presented with a series of trouble tickets related to issues introduced during these configurations.
Note: Although trouble tickets have many similar fault indications, each ticket has its own issue and solution.
Each ticket has 3 sub questions that need to be answered & topology remains same.
Question-1 Fault is found on which device,
Question-2 Fault condition is related to,
Question-3 What exact problem is seen & what needs to be done for solution


Client is unable to ping IP 209.65.200.241
Solution
Steps need to follow as below:-
* When we check on client 1 & Client 2 desktop we are not receiving DHCP address from R4 Ipconfig ----- Client will be getting 169.X.X.X
* On ASW1 port Fa1/0/ 1 & Fa1/0/2 access port VLAN 10 was assigned which is using IP address 10.2.1.0/24 Sh run ------- & check for running config of int fa1/0/1 & fa1/0/2
interface FastEthernet1/0/1
switchport mode access
switchport access vlan 10
interface FastEthernet1/0/2
switchport mode access
switchport access vlan 10

Change required:
On ASW1 below change is required for switch-to-switch connectivity.
int range portchannel13,portchannel23
switchport trunk allowed vlan none
switchport trunk allowed vlan 10,200

NEW QUESTION: 3
Given: What is result?

A. Successful
B. Compilation fails
C. An exception is thrown at runtime
D. Unsuccessful
Answer: B

NEW QUESTION: 4
Given this enum declaration:

Examine this code:
System.out.println(Letter.values()[1]);
What code should be written at line 5 for this code to print 200?
A. public String toString() { return String.valueOf(Letter.values()[1]); }
B. public String toString() { return String.valueOf(ALPHA.v); }
C. String toString() { return "200"; }
D. public String toString() { return String.valueOf(v); }
Answer: D
Explanation: