H35-460 Practice Questions

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NEW QUESTION: 1

A. Set-ADFSProperties -ProxyTrustTokenLifetime 1:00:00
B. Set-ADFSProperties -ExtendedProtectionTokenCheck None
C. Set-ADFSProperties -SSOLifetime 1:00:00
D. Set-ADFSProperties -AddProxyAuthenticationRules None
Answer: B
Explanation:
Explanation
Explanation/Reference:
Certain client browser software, such as Firefox, Chrome, and Safari, do not support the Extended Protection for Authentication capabilities that can be used across the Windows platform to protect against man-in-the-middle attacks. To prevent this type of attack from occurring over secure AD FS communications, AD FS 2.0 enforces (by default) that all communications use a channel binding token (CBT) to mitigate against this threat.
Note: Disable the extended Protection for authentication
To disable the Extended Protection for Authentication feature in AD FS 2.0
* On a federation server, login using the Administrator account, open the Windows PowerShell command prompt, and then type the following command:
Set-ADFSProperties -ExtendedProtectionTokenCheck None
* Repeat this step on each federation server in the farm.
Reference: Configuring Advanced Options for AD FS 2.0

NEW QUESTION: 2
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.1.0.0.0.254
B. 許可10.0.0.0.255.255.255.254
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.0.0.0.1
Answer: A
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s 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 0.0.15.255 172.23.31.255.
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 000 ) 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 oddnumbered hosts in the 10.0.0.0/24 subnet.

NEW QUESTION: 3
You are designing a logo that needs to look smooth when resized larger or smaller. Which application should you use?
A. Adobe Photoshop
B. Adobe Streamline
C. Adobe Frame Maker
D. AdobeIllustrator
Answer: D

NEW QUESTION: 4
ワイヤレスネットワークに接続されたラップトップを使用している従業員。インターネットには接続できますが、社内リソースには接続できません。
システム管理者は、ラップトップから次の情報を確認します。

管理者は、10.100.0.0 / 23内のネットワークアドレスが表示されることを期待していました。これは会社のワイヤレスネットワークと一致しています。ネットワーク管理者は、会社のワイヤレスネットワークが構成され、正しく機能していることを確認します。この問題の最も可能性の高い原因は次のうちどれですか？
A. セキュリティタイプの不一致
B. ハンネルオーバーラップ
C. 干渉
D. 間違ったSSID
Answer: D