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NEW QUESTION: 1
Note: This question is part of a series of questions that present the same scenario. Each question
in the series contains a unique solution that might meet the stated goals. Some question sets
might have more than one correct solution, while others might not have a correct solution.
After you answer a question in this section, you will NOT be able to return to it. As a result, these
questions will not appear in the review screen.
You have an Azure Subscription named Sub1.
You have an Azure Storage account named Sa1 in a resource group named RG1.
Users and applications access the blob service and the file service in Sa1 by using several shared access
signatures (SASs) and stored access policies.
You discover that unauthorized users accessed both the file service and the blob service.
You need to revoke all access to Sa1.
Solution: You create a new stored access policy.
Does this meet the goal?
A. Yes
B. No
Answer: A
Explanation:
Explanation/Reference:
Explanation:
To revoke a stored access policy, you can either delete it, or rename it by changing the signed identifier.
Changing the signed identifier breaks the associations between any existing signatures and the stored
access policy. Deleting or renaming the stored access policy immediately effects all of the shared access
signatures associated with it.
References:
https://docs.microsoft.com/en-us/rest/api/storageservices/Establishing-a-Stored-Access-Policy

NEW QUESTION: 2
What would be a test approach regarding the test design techniques to be applied that would fit an item with the highest risk? 2 credits
A. Component testing: statement testing; System testing: equivalence partitioning
B. Component testing: decision testing; System testing: decision table testing
C. Component testing: decision testing; System testing: exploratory testing
D. Component testing: statement testing; System testing: decision table partitioning
Answer: B

NEW QUESTION: 3
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.255.255.255.254
B. 許可10.0.0.1.0.0.0.0
C. 許可10.0.0.0.0.0.0.1
D. 許可10.0.0.1.0.0.0.254
Answer: D
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

A. vSphere Standard license
B. 10 GbE uplink for FT logging network
C. 1 GbE uplink for FT logging network
D. vSphere Enterprise Plus license
Answer: B,D
Explanation:
Explanation: References: