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NEW QUESTION: 1
Note: This question is part of a series of questions that use the same scenario. For your convenience, the scenario is repeated in each question. Each question presents a different goal and answer choices, but the text of the scenario is the same in each question in this series.
Start of repeated scenario
You have six workbook queries that each extracts a table from a Microsoft Azure SQL database. The table are loaded to the data model, but the data is not loaded to any worksheets. The data model is shown in the Data Model exhibit. (Click the Exhibit button.) Your company has 100 product subcategories and more than 10,000 products.
End of repeated scenario.
You have a PivotChart that uses Manufacturer as the axis and the sum of SalesAmount as the values.
You need to ensure that only the top 10 manufactures appear in the chart.
What should you do?
A. Configure the Value Filters.
B. Create a calculated column.
C. Change the format of the SalesAmount field.
D. Summarize the SaleAmount field by Max.
Answer: B
NEW QUESTION: 2
A solution which tracks and organizes the usage and cost of voice and data is referred to as:
A. SaaS.
B. TEM.
C. B2B.
D. MaaS.
Answer: B
NEW QUESTION: 3
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 10.0.0.0.0.0.0.1
C. Permit 10.0.0.1.0.0.0.254
D. Permit 10.0.0.0.255.255.255.254
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.
NEW QUESTION: 4
You are configuring a Windows Server 2016 failover cluster in a workgroup.
Before installing one of the nodes, you run the ipconfig /all command and receive the following Output
You need to ensure that Server 1 can be added as a node in the cluster What should you do?
A. Configure a DNS suffix
B. Assign a static IP address
C. Change the Node Type to Broadcast
D. Enable NetBIOS over TCP/IP
Answer: A