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NEW QUESTION: 1
仮想マシン環境でブロードキャスト放射線が発生する2つの理由は何ですか？ （2つ選択してください。）
A. レイヤー2ドメインは、仮想マシン環境で大きくなる可能性があります。
B. vSwitchとネットワークスイッチ間の通信はブロードキャストベースです。
C. ブロードキャストパケットを処理するには、vSwitchがサーバーのCPUに割り込む必要があります。
D. vSwitchとネットワークスイッチ間の通信はマルチキャストベースです。
E. 仮想マシンは主にブロードキャストモードで通信します。
Answer: A,E
Explanation:
Explanation
Broadcast radiation is the accumulation of broadcast and multicast traffic on a computer network. Extreme amounts of broadcast traffic constitute a broadcast storm.
The amount of broadcast traffic you should see within a broadcast domain is directly proportional to the size of the broadcast domain. Therefore if the layer 2 domain in virtual machine environment is too large, broadcast radiation may occur -> VLANs should be used to reduce broadcast radiation.
Also if virtual machines communicate via broadcast too much, broadcast
radiation may occur.
Another reason for broadcast radiation is using a trunk (to extend VLANs) from the network switch to the physical server.
Note about the structure of virtualization in a hypervisor:
Hypervisors provide virtual switch (vSwitch) that Virtual Machines (VMs) use to communicate with other VMs on the same host. The vSwitch may also be connected to the host's physical NIC to allow VMs to get layer 2 access to the outside world.
Each VM is provided with a virtual NIC (vNIC) that is connected to the
virtual switch. Multiple vNICs can connect to a single vSwitch, allowing VMs on a physical host to communicate with one another at layer 2 without having to go out to a physical switch.

Although vSwitch does not run Spanning-tree protocol but vSwitch
implements other loop prevention mechanisms. For example, a
frame that enters from one VMNIC is not going to go out of the
physical host from a different VMNIC card.

NEW QUESTION: 2
ワイヤレスネットワークは、コントローラーベースのAPとバージョン7.0のMR1を使用しています。 APはコントローラに正しく接続されているようです。 APの1つ近くの無線ユーザが、事前共有キーを使用してネットワークに接続できないことを報告します。どのオプションが、そのクライアントの接続状態を確認できるGUIパスを表示しますか？
A. 無線 > 802.11b/g/n > クライアント
B. モニター > クライアント
C. セキュリティ> Advanced > クライアント
D. セキュリティ>クライアント
E. WLAN > WLAN_ID > 進められます。
Answer: B

NEW QUESTION: 3
Given:
class Base {
// insert code here
}
public class Derived extends Base{
public static void main(String[] args) {
Derived obj = new Derived();
obj.setNum(3);
System.out.println("Square = " + obj.getNum() * obj.getNum());
}
}
Which two options, when inserted independently inside class Base, ensure that the class is being properly encapsulated and allow the program to execute and print the square of the number?
A. protected int num; private int getNum() { return num; } public void setNum(int num) { this.num = num;}
B. private int num;public int getNum() {return num;} private void setNum(int num) { this.num = num;}
C. private int num; public int getNum() { return num; }public void setNum(int num) { this.num = num;}
D. protected int num; public int getNum() { return num; } public void setNum(int num) { this.num = num;}
E. public int num; protected public int getNum() { return num; }protected public void setNum(int num) { this.num = num;}
Answer: C,D
Explanation:
Incorrect:
Not B: illegal combination of modifiers: protected and public
not C: setNum method cannot be private. not E: getNum method cannot be private.

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