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NEW QUESTION: 1
君は
ビデオオンデマンドストリーミングメディアを配信するために、Azure App ServiceがホストするASP.NET Core Webアプリケーションを開発しています。 Webエンドポイントに対してAzureコンテンツ配信ネットワーク（CDN）規格を有効にします。顧客のビデオは、次のURLの例を使用してWebアプリケーションからダウンロードされます。
http://www.contoso.com/content.mp4?quality=l
すべてのメディアコンテンツは1時間後にキャッシュから期限切れになる必要があります。品質が異なるカスタマービデオは、最も近い地域のPOP（Point of Presence）ノードに配信する必要があります。
Azure CDNキャッシュルールを構成する必要があります。
どのオプションを使うべきですか？回答するには、回答領域で適切なオプションを選択します。
注：それぞれ正しい選択は1ポイントの価値があります。
Answer:
Explanation:
NEW QUESTION: 2
You are customizing standard content types for an organization. For the tracking of a department, you must allow users to easily select or enter a value.
In some cases, users may refer to a single department by more than one name. Users must be able to select or type any name for a department but the document must be associated with only one department.
You need to create a column that supports the requirements.
What should you do?
A. Create a Choice list that uses a drop-downmenu.
B. Create a Managed Metadata site column.
C. Create a Choice site column that uses a drop-down menu.
D. Create a Choice site column that uses check boxes.
Answer: C
Explanation:
http://en.share-gate.com/blog/start-learn-sharepoint-advanced-column-types
NEW QUESTION: 3
Why would you use a virtual master data types?
A. To combine two or more independent simple master data types
B. To reference an existing master data type
C. To references a master data from an external source
D. To create a join between two or more master data types
Answer: D
NEW QUESTION: 4
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.0
B. Permit 10.0.0.0.255.255.255.254
C. Permit 10.0.0.1.0.0.0.254
D. Permit 10.0.0.0.0.0.0.1
Answer: C
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 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 oddnumbered hosts in the 10.0.0.0/24 subnet.