TDA-C01 Exam Registration & Exam TDA-C01 Braindumps - TDA-C01 Visual Cert Exam - Sapsam

  • Exam Code: TDA-C01
  • Exam Name: Tableau Certified Data Analyst
  • Version: V15.35     Q & A: 208 Questions and Answers

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NEW QUESTION: 1
有効された自動セグメント領域管理(ASSM)とローカル管理表領域に格納されている販売テーブルの構造を調べています。

最高水位標の下の断片化した空き領域を再利用するために、オンラインによるセグメントの縮小を実行します。
操作の開始前に何を確認する必要がありますか。
A. 行の移動が有効になっています。
B. テーブルの参照整合性制約が無効になっています。
C. 保留中のトランザクションがテーブルに存在しません。
D. このテーブルの上に実行されているクエリはありません。
E. セグメントのサイズに余分なディスクスペースと同等の表領域で使用可能です。
Answer: A

NEW QUESTION: 2
Which of the following is a logical boundary for vMotion?
A. Cluster
B. Resource Pool
C. Folder
D. vDC
Answer: D
Explanation:
By design, vMotion is possible from a logical perspective between the vDC. It is possible, however, to have a stretched
cluster and to
not
satisfy
vMotion requirements. The logical boundary is limited to vDCs.

NEW QUESTION: 3
SIMULATION





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: 4
Refer to the exhibit.

How should a Synergy integrator pair interconnect bays to set up redundant fabrics in an HPE Synergy 12000 Frame?
A. Interconnect bays 1 and 1 and 6: bays 2 and 5: bays 3 and 4.
B. Interconnect 1 bays 1 and 4: bays 2 and 5: bays 3 and 6.
C. Interconnect bays 1 and 3: bays 2 and 4: bays 5 and 6.
D. Interconnect bays 1 and bays 2: bays 3 and 4: bays 5 and 6.
Answer: A

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