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- Name: HPE Hybrid Cloud Solutions
- Exam Code: HPE0-V25
- Certification: HPE0-V25
- Vendor: HP
- Total Question: 630
- Price: 49$
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NEW QUESTION: 1
You have a computer that runs Windows XP Service Pack 2 (SP2).
You need to upgrade the operating system to Windows 7.
You must achieve this goal in the minimum amount of time.
What should you do?
A. Upgrade to Windows Vista SP2. From the Windows 7 installation media, run Setup.exe and select the Custom (advanced) option.
B. Upgrade to Windows Vista SP2. From the Windows 7 installation media, run Setup.exe and select the Upgrade option.
C. Install Windows XP Service Pack 3 (SP3). Run Setup.exe from the Windows 7 installation media and select the Custom (advanced) option.
D. Install Windows XP Service Pack 3 (SP3). Run Setup.exe from the Windows 7 installation media and select the Upgrade option.
Answer: B
Explanation:
Explanation/Reference:
10278 40176
NEW QUESTION: 2
A network associate is adding security to the configuration of the Corp1 router. The user on host C should be able to use a web browser to access financial information from the Finance Web Server. No other hosts from the LAN nor the Core should be able to use a web browser to access this server. Since there are multiple resources for the corporation at this location including other resources on the Finance Web Server, all other traffic should be allowed.
The task is to create and apply an access-list with no more than three statements that will allow ONLY host C web access to the Finance Web Server. No other hosts will have web access to the Finance Web Server. All other traffic is permitted.
Access to the router CLI can be gained by clicking on the appropriate host.
All passwords have been temporarily set to "cisco".
The Core connection uses an IP address of 198.18.196.65.
The computers in the Hosts LAN have been assigned addresses of 192.168.33.1 - 192.168.33.254 host A 192.168.33.1 host B 192.168.33.2 host C 192.168.33.3 host D 192.168.33.4 The servers in the Server LAN have been assigned addresses of 172.22.242.17 - 172.22.242.30.
The Finance Web Server is assigned an IP address of 172.22.242.23.
A. Configuring ACL
Corp1>enable
Corp1#configure terminal
comment: To permit only Host C (192.168.33.3){source addr} to access finance server address (172.22.242.23) {destination addr} on port number 80 (web) Corp1(config)#access-list 100 permit tcp host 192.168.33.3 host 172.22.242.23 eq 80 comment: To deny any source to access finance server address (172.22.242.23) {destination addr} on port number 80 (web) Corp1(config)#access-list 100 deny tcp any host 172.22.242.23 eq 80 comment: To permit ip protocol from any source to access any destination because of the implicit deny any any statement at the end of ACL.
Corp1(config)#access-list 100 permit ip any any
Applying the ACL on the Interface
comment: Check show ip interface brief command to identify the interface type and number by checking the IP address configured.
Corp1(config)#interface fa 0/1
Step 3: Only Host C (192.168.33.3) has access to the server. If the other host can also access then maybe something went wrong in your configuration. Check whether you configured correctly and in order.
Step 4: If only Host C (192.168.33.3) can access the Finance Web Server you can click on NEXT button to successfully submit the ACL SIM.
B. Configuring ACL
Corp1>enable
Corp1#configure terminal
comment: To permit only Host C (192.168.33.3){source addr} to access finance server address (172.22.242.23) {destination addr} on port number 80 (web) Corp1(config)#access-list 100 permit tcp host 192.168.33.3 host 172.22.242.23 eq 80 comment: To deny any source to access finance server address (172.22.242.23) {destination addr} on port number 80 (web) Corp1(config)#access-list 100 deny tcp any host 172.22.242.23 eq 80 comment: To permit ip protocol from any source to access any destination because of the implicit deny any any statement at the end of ACL.
Corp1(config)#access-list 100 permit ip any any
Applying the ACL on the Interface
comment: Check show ip interface brief command to identify the interface type and number by checking the IP address configured.
Corp1(config)#interface fa 0/1
If the ip address configured already is incorrect as well as the subnet mask. This should be corrected in order ACL to work type this commands at interface mode :
no ip address 192.x.x.x 255.x.x.x (removes incorrect configured ipaddress and subnet mask) Configure Correct IP Address and subnet mask:
ip address 172.22.242.30 255.255.255.240 ( range of address specified going to server is given as
172.22.242.17 - 172.22.242.30 )
Comment: Place the ACL to check for packets going outside the interface towards the finance web server.
Corp1(config-if)#ip access-group 100 out
Corp1(config-if)#end
Important: To save your running config to startup before exit.
Corp1#copy running-config startup-config
Verifying the Configuration:
Step1: show ip interface brief command identifies the interface on which to apply access list.
Step2: Click on each host A, B, C, & D.
Host opens a web browser page, Select address box of the web browser and type the ip address of finance web server (172.22.242.23) to test whether it permits /deny access to the finance web Server.
Step 3: Only Host C (192.168.33.3) has access to the server. If the other host can also access then maybe something went wrong in your configuration. Check whether you configured correctly and in order.
Step 4: If only Host C (192.168.33.3) can access the Finance Web Server you can click on NEXT button to successfully submit the ACL SIM.
Answer: B
NEW QUESTION: 3
Drag and Drop Question
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right Select and Place:
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.
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