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NEW QUESTION: 1
A company has contracted with a third party to audit several AWS accounts. To enable the audit, cross-account IAM roles have been created in each account targeted for audit. The Auditor is having trouble accessing some of the accounts.
Which of the following may be causing this problem? (Choose three.)
A. The external ID used by the Auditor is missing or incorrect.
B. The secret key used by the Auditor is missing or incorrect.
C. The Auditor is using the incorrect password.
D. The Amazon EC2 role used by the Auditor must be set to the destination account role.
E. The role ARN used by the Auditor is missing or incorrect.
F. The Auditor has not been granted sts:AssumeRole for the role in the destination account.
Answer: A,E,F

NEW QUESTION: 2
You have an on-premises data center and an Azure subscription. The data center contains two VPN devices.
The subscription contains an Azure virtual network named VNet1. VNet1 contains a gateway subnet.
You need to create a site-to-site VPN. The solution must ensure that is a single instance of an Azure VPN gateway fails, or a single on-premises VPN device fails, the failure will not cause an interruption that is longer than two minutes.
What is the minimum number of public IP addresses, virtual network gateways, and local network gateways required in Azure? To answer, select the appropriate options in the answer area.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation

Box 1: 4
Two public IP addresses in the on-premises data center, and two public IP addresses in the VNET.
The most reliable option is to combine the active-active gateways on both your network and Azure, as shown in the diagram below.

Box 2: 2
Every Azure VPN gateway consists of two instances in an active-standby configuration. For any planned maintenance or unplanned disruption that happens to the active instance, the standby instance would take over (failover) automatically, and resume the S2S VPN or VNet-to-VNet connections.
Box 3: 2
Dual-redundancy: active-active VPN gateways for both Azure and on-premises networks References:
https://docs.microsoft.com/en-us/azure/vpn-gateway/vpn-gateway-highlyavailable

NEW QUESTION: 3
You have an Azure subscription.
You plan to deploy two Azure web apps that have the requirements shown in the following table.

You need to select the App Service plans for the web apps. The solution must minimize costs.
Which App Service plan should you select for each web app? To answer, select the appropriate options in the answer area.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation


Reference:
https://azure.microsoft.com/en-us/pricing/details/app-service/plans/

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.