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NEW QUESTION: 1
Within Cisco Prime Infrastructure, which configuration Archive task will allow you to specify when to copy the running configuration to the startup configuration?
A. Schedule Deploy
B. Schedule Rollback
C. Schedule Overwrite
D. Schedule Archive
Answer: C
Explanation:
You can schedule to have Prime Infrastructure copy the running configuration to the startup configuration by choosing Inventory > Device Configuration Archive, then clicking Schedule Overwrite .
http://www.cisco.com/c/en/us/td/docs/net_mgmt/prime/infrastructure/3-0/user/guide/pi_ug/ chgdevconfig.html#82530

NEW QUESTION: 2
モデルの適合の問題を修正する必要があります。
順番に実行する必要がある3つのアクションはどれですか？回答するには、適切なアクションをアクションのリストから回答エリアに移動し、正しい順序に並べます。

Answer:
Explanation:

Explanation

Step 1: Augment the data
Scenario: Columns in each dataset contain missing and null values. The datasets also contain many outliers.
Step 2: Add the Bayesian Linear Regression module.
Scenario: You produce a regression model to predict property prices by using the Linear Regression and Bayesian Linear Regression modules.
Step 3: Configure the regularization weight.
Regularization typically is used to avoid overfitting. For example, in L2 regularization weight, type the value to use as the weight for L2 regularization. We recommend that you use a non-zero value to avoid overfitting.
Scenario:
Model fit: The model shows signs of overfitting. You need to produce a more refined regression model that reduces the overfitting.

NEW QUESTION: 3
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選択して配置できる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.