Answer:
Question is incomplete. it needs a topology diagram and also it needs Router R1 table. I assume User has access to the topology and Routing table.
Below Configuration will help to fix ACL problem
Hosts from the 172.16.0.0/16 network should have full access to Server1, Server2 and Server3 but this is not currently the case, as L1 can’t communicate to Server2 or Server3.
Explanation:
Following Configuration on Cisco Router R1 will help to fix all the ACL problems.
enable
configure terminal
no ip access-list standard FROM_10
ip access-list standard FROM_10
deny host 10.0.0.2
permit any
exit
!
no ip access-list standard FROM_172
ip access-list standard FROM_172
permit host 172.16.0.2
exit
!
interface GigabitEthernet0/0
ip access-group FROM_192 out
end
write memory
!
Answer:
The correct answer to the following question will be "Option 4".
Explanation:
Group seems to have a backlog optimization conversation with either the company owner after each iteration to obtain insight about the backlog things to have been collected in the future iterations or executions.
- Without the need to get feedback on the backlog or pending things in preparation for potential implementations.
- The next phase brings the dedicated backlog to completion. Iteration targets are calculated, based on dedicated research.
Answer:
Yo creo que él internet a mejorado nuestra vida pero también nos a embobado porque lo que hacemos mas es buscar cosas que no aportan a tu vida y no cosas que nos ayuden en la vida.
Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Answer:
Object-Oriented DBMS.
Explanation:
Like classes in Object oriented programming Object-Oriented DBMS is can store objects.It follows an Object oriented data model with the properties of classes and the properties of Object oriented programming also.Objects can be in the form of complex data types or different data types.
