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弊社のRoyalholidayclubbedはIT認定試験のソフトの一番信頼たるバンドになるという目標を達成するために、弊社はあなたに最新版のCiscoの350-401日本語版問題集試験問題集を提供いたします。弊社のソフトを使用して、ほとんどのお客様は難しいと思われているCiscoの350-401日本語版問題集試験に順調に剛角しました。これも弊社が自信的にあなたに商品を薦める原因です。 Royalholidayclubbed350-401日本語版問題集問題集は試験の範囲を広くカバーするだけでなく、質は高いです。Royalholidayclubbedの350-401日本語版問題集問題集を購入し勉強するだけ、あなたは試験にたやすく合格できます。 これをよくできるために、我々は全日24時間のサービスを提供します。

それはRoyalholidayclubbedの350-401日本語版問題集問題集です。

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全力を尽くせば、Cisco 350-401日本語版問題集試験の合格も可能となります。

Royalholidayclubbedはきみの貴重な時間を節約するだけでなく、 安心で順調に試験に合格するのを保証します。Royalholidayclubbedは専門のIT業界での評判が高くて、あなたがインターネットでRoyalholidayclubbedの部分のCisco 350-401日本語版問題集「Implementing Cisco Enterprise Network Core Technologies (350-401 ENCOR)」資料を無料でダウンロードして、弊社の正確率を確認してください。弊社の商品が好きなのは弊社のたのしいです。

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350-401 PDF DEMO:

QUESTION NO: 1
What is the correct EBGP path attribute list, ordered from most preferred to the least preferred, that the BGP best-path algorithm uses?
A. weight, local preference, AS path, MED
B. weight, AS path, local preference, MED
C. local preference, weight MED, AS path
D. local preference, weight, AS path, MED
Answer: A
Explanation:
Path Selection Attributes: Weight > Local Preference > Originate > AS Path > Origin > MED > External
> IGP Cost > eBGP Peering > Router ID

QUESTION NO: 2
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.255.255.255.254
B. Permit 10.0.0.1.0.0.0.254
C. Permit 10.0.0.1.0.0.0.0
D. Permit 10.0.0.0.0.0.0.1
Answer: B
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum:
0001111 = 31).
The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a
"0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all odd- numbered hosts in the 10.0.0.0/24 subnet.

QUESTION NO: 3
Which access controls list allows only TCP traffic with a destination port range of 22-433, excluding port 80?
A. Deny tcp any any ne 80
Permit tcp any any range 22 443
B. Permit tcp any any ne 80
C. Permit tco any any range 22 443
Deny tcp any any eq 80
D. Deny tcp any any eq 80
Permit tco any any gt 21 it 444
Answer: D

QUESTION NO: 4
Which statement about multicast RPs is true?
A. RPs are required for protocol independent multicast sparse mode and dense mode.
B. By default, the RP is needed only to start new sessions with sources and receivers.
C. By default, the RP is needed periodically to maintain sessions with sources and receivers
D. RPs are required only when using protocol independent multicast dense mode.
Answer: B
Explanation:
A rendezvous point (RP) is required only in networks running Protocol Independent Multicast sparse mode (PIM-SM).
By default, the RP is needed only to start new sessions with sources and receivers.
Reference:
https://www.cisco.com/c/en/us/td/docs/ios/solutions_docs/ip_multicast/White_papers/rps.html
For your information, in PIM-SM, only network segments with active receivers that have explicitly requested multicast data will be forwarded the traffic. This method of delivering multicast data is in contrast to the PIM dense mode (PIM-DM) model. In PIM-DM, multicast traffic is initially flooded to all segments of the network. Routers that have no downstream neighbors or directly connected receivers prune back the unwanted traffic.

QUESTION NO: 5
Which statement explains why Type 1 hypervisor is considered more efficient than Type 2 hypervisor?
A. Type 1 hypervisor enables other operating systems to run on it
B. Type 1 hypervisor runs directly on the physical hardware of the host machine without relying on the underlying OS
C. Type 1 hypervisor relics on the existing OS of the host machine to access CPU, memory, storage, and network resources.
D. Type 1 hypervisor is the only type of hypervisor that supports hardware acceleration techniques
Answer: B
Explanation:
There are two types of hypervisors: type 1 and type 2 hypervisor.
In type 1 hypervisor (or native hypervisor), the hypervisor is installed directly on the physical server.
Then instances of an operating system (OS) are installed on the hypervisor. Type 1 hypervisor has direct access to the hardware resources. Therefore they are more efficient than hosted architectures.
Some examples of type 1 hypervisor are VMware vSphere/ESXi, Oracle VM Server, KVM and
Microsoft Hyper-V.
In contrast to type 1 hypervisor, a type 2 hypervisor (or hosted hypervisor) runs on top of an operating system and not the physical hardware directly. A big advantage of Type 2 hypervisors is that management console software is not required.

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Updated: May 28, 2022

 

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