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NEW QUESTION: 1
A particular database threat utilizes a SQL injection technique to penetrate a target system. How would an attacker use this technique to compromise a database?
A. An attacker uses poorly designed input validation routines to create or alter SQL commands to gain access to unintended data or execute commands of the database
B. An attacker submits user input that executes an operating system command to compromise a target system
C. An attacker gains control of system to flood the target system with requests,preventing legitimate users from gaining access
D. An attacker utilizes an incorrect configuration that leads to access with higher-thanexpected privilege of the database
Answer: A
Explanation:
Using the poorly designed input validation to alter or steal data from a database is a SQL injection attack.
NEW QUESTION: 2
A. Option B
B. Option A
C. Option C
D. Option D
Answer: D
Explanation:
Explanation
On DSW1, related to HSRP, under VLAN 10 change the given track 1 command to instead use the track 10 command.
Topic 13, Ticket 13: DHCP Issue
Topology Overview (Actual Troubleshooting lab design is for below network design)
* Client Should have IP 10.2.1.3
* EIGRP 100 is running between switch DSW1 & DSW2
* OSPF (Process ID 1) is running between R1, R2, R3, R4
* Network of OSPF is redistributed in EIGRP
* BGP 65001 is configured on R1 with Webserver cloud AS 65002
* HSRP is running between DSW1 & DSW2 Switches
The company has created the test bed shown in the layer 2 and layer 3 topology exhibits.
This network consists of four routers, two layer 3 switches and two layer 2 switches.
In the IPv4 layer 3 topology, R1, R2, R3, and R4 are running OSPF with an OSPF process number 1.
DSW1, DSW2 and R4 are running EIGRP with an AS of 10. Redistribution is enabled where necessary.
R1 is running a BGP AS with a number of 65001. This AS has an eBGP connection to AS 65002 in the ISP's network. Because the company's address space is in the private range.
R1 is also providing NAT translations between the inside (10.1.0.0/16 & 10.2.0.0/16) networks and outside (209.65.0.0/24) network.
ASW1 and ASW2 are layer 2 switches.
NTP is enabled on all devices with 209.65.200.226 serving as the master clock source.
The client workstations receive their IP address and default gateway via R4's DHCP server.
The default gateway address of 10.2.1.254 is the IP address of HSRP group 10 which is running on DSW1 and DSW2.
In the IPv6 layer 3 topology R1, R2, and R3 are running OSPFv3 with an OSPF process number 6.
DSW1, DSW2 and R4 are running RIPng process name RIP_ZONE.
The two IPv6 routing domains, OSPF 6 and RIPng are connected via GRE tunnel running over the underlying IPv4 OSPF domain. Redistrution is enabled where necessary.
Recently the implementation group has been using the test bed to do a 'proof-of-concept' on several implementations. This involved changing the configuration on one or more of the devices. You will be presented with a series of trouble tickets related to issues introduced during these configurations.
Note: Although trouble tickets have many similar fault indications, each ticket has its own issue and solution.
Each ticket has 3 sub questions that need to be answered & topology remains same.
Question-1 Fault is found on which device,
Question-2 Fault condition is related to,
Question-3 What exact problem is seen & what needs to be done for solution
Solution
Steps need to follow as below:-
* When we check on client 1 & Client 2 desktop we are not receiving DHCP address from R4 ipconfig ----- Client will be receiving Private IP address 169.254.X.X
* From ASW1 we can ping 10.2.1.254....
* On ASW1 VLAN10 is allowed in trunk & access command will is enabled on interface but DHCP IP address is not recd.
On R4 the DHCP IP address is not allowed for network 10.2.1.0/24 which clearly shows the problem lies on R4 & the problem is with DHCP
NEW QUESTION: 3
A project requires an initial outlay of 650. It also needs capital spending of 700 at the end of year 1 and
900 at the end of year 2. It has no revenues for the first 2 years but receives 1,200 in year 3, 1,600 in year
4 and 2,300 in year 5. The project's payback period equals ________.
A. 2.26 years
B. 4.54 years
C. 3.66 years
D. 4.91 years
Answer: D
Explanation:
Explanation/Reference:
Explanation:
The cash flows of the project starting at the end of year 1 are:
-700, -900, +1,200, +1,600, +2,300
The payback period is defined as the expected number of years that would be required to recover the original investment. In particular, Payback period = Years before full recovery + (unrecovered cost at the start of payback year)/(net cash flow in the payback year) The net account balance goes positive in the 4th year. At the beginning of the 4th year, the outstanding balance equals 650+700+900-1,200 = $1,050.
Therefore, payback period = 3 + 1,050/1,600 = 3.66 years.
NEW QUESTION: 4
A fiber company has acquired permission to bury a fiber cable through a famer's land. Which of the following should be in the agreement with the farmer to protect the availability of the network?
A. No crops will be planted on top of the cable
B. No farm animals will graze near the burial site of the cable
C. No digging will occur near the burial site of the cable
D. No buildings or structures will be placed on top of the cable
Answer: C