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NEW QUESTION: 1
Azure ExpressRoute回線を構成する必要があります。
Azure ExpressRouteルーティングをどのように構成する必要がありますか？答えるには、適切な構成を正しい場所にドラッグします。各構成は、1回、複数回、またはまったく使用できません。コンテンツを表示するには、ペイン間で分割バーをドラッグするか、スクロールする必要がある場合があります。
注：それぞれの正しい選択には1ポイントの価値があります。
Answer:
Explanation:
NEW QUESTION: 2
プロジェクトマネージャーがリソースの平準化を実行していて、2番目のプロジェクトで機能テスターの1人に40％が割り当てられていることに気付きました。プロジェクトマネージャーは、リソースを100％プロジェクト専用にする必要があります。そうしないと、テストフェーズで納期が危険にさらされます。プロジェクトマネージャーが次に行うべきことは次のうちどれですか？
A. リソースマネージャーに、割り当ての40％で新しいリソースを割り当てるように依頼します。
B. リソースを割り当ての60％に変更し、緩和計画を確認します。
C. 当初の計画に従って、プロジェクトでリソースを100％割り当てたままにします。
D. 2番目のプロジェクトからリソースの割り当てを解除するようにリソースマネージャーに依頼します。
Answer: B
NEW QUESTION: 3
HOTSPOT
Answer:
Explanation:
NEW QUESTION: 4
質問をドラッグアンドドロップ
IPv4ネットワークサブネットを左側から右側の正しい使用可能なホスト範囲にドラッグアンドドロップします
Answer:
Explanation:
Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet 172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111
1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is
172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or
172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet.
Therefore the last usable host is 172.28.229.254.
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