What hardware components are generally required for a desktop computing system?

Dennis is driving a car with his family onboard. His children are sitting on the backseat. They have a tablet and want to watch

Mice21 [21]

Answer:

A and D could work

Explanation:

they do basically does the same thing!!

8 0

4 years ago

Different network scenarios require the

Vladimir [108]

Answer:

a) Viruses

Explanation:

Vulnerability assessment is performed by variety of tools and these can be protocol analyzer, vulnerability scanner, Honeypots, port scanners, honey nets and banner grabbing tools.

Port scanners : they are used to scan the ports which can be exploited by attackers. Most of TCP/IP applications communicate using different ports and attackers can exploit these ports if they are found vulnerable. for example : ports like 80 and 443 are commonly used for HTTP and SSL communication. 22 is used for SSH and if they are open to world then it will allow attackers to use those to get the entry into the system.

Protocol Analyzers : They are used to analyze the packet captures. tools like wireshark, tshark, these tools will help user to decode HTTP/SSL/HTTPS/FTP/RSTP or any application protocols communication. this will help user to understand any unwanted or non-anticipated traffic.

Vulnerability Scanners : They are used to detect the vulnerabilities in the network or systems. these will help administrators to get the alerts whenever there are unanticipated activity. It will have two types of scanners one is active scanner and other is passive scanner. Active scanners will keep on sending probes at the fixed time slots and passibe scanners will be in listening mode all the time. These scanners will alert when a new nodes comes up or goes down or if any system gets compromised.

Honeypots and Honey nets : They are devices or softwares which are having limited security. These devices are made vulnerable purposefully so that attackers can try and attack. This will help softwares to detect the attackers when they try to exploit these open vulnerabilities. They are deceptions created with purpose.

Banner grabbing Tools : These are tools which will capture the banner information like HTTP protocol version, underlying operating system, open ssl versions being used, server software and similar information. these are important as it will open up potential vulnerabilities in the underlying software. for ex : SSLv3 has a vulnerability known to world and can be exploited by anyone.

4 0

3 years ago

Define an organizational unit in your own words.

Musya8 [376]

Explanation:

An organizational unit (OU) is a subdivision within an Active Directory (hierarchical collection of network resources that can contain users, computers, printers, and other Active Directories) into which you can place users, groups, computers, and other organizational units. Each domain can implement its own organizational unit hierarchy, typically used either to differentiate between objects with the same name, or to parcel out authority to create and manage objects.

I hope you find this information useful and interesting! Good luck!

3 0

4 years ago

Write a program using integers userNum and x as input, and output userNum divided by x four times. Ex: If the input is: 2000 2

STatiana [176]

Answer:

// here is code in c++.

// include header

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variables to read input

int userNum,x;

cout<<"enter the value of userNum and x :";

// read the input from user

cin>>userNum>>x;

// divide the userNum with x 4 times

for(int a=0;a<4;a++)

{

userNum=userNum/x;

cout<<userNum<<" ";

}

cout<<endl;

return 0;

}

Explanation:

Declare two variables "userNum" and "x". Read the value of these. Run a for loop 4 time and divide the "userNum" with "x" and print the value of "userNum".

<u>Output:</u>

enter the value of userNum and x :2000 2

1000 500 250 125

3 0

3 years ago

Determine and prove whether an argument in English is valid or invalid. About Prove whether each argument is valid or invalid. F

yawa3891 [41]

Answer:

Each understudy on the respect roll got an A.

No understudy who got a confinement got an A.

No understudy who got a confinement is on the respect roll.

No understudy who got an A missed class.

No understudy who got a confinement got an A.

No understudy who got a confinement missed class

Explanation:

M(x): x missed class

An (x): x got an A.

D(x): x got a confinement.

¬∃x (A(x) ∧ M(x))

¬∃x (D(x) ∧ A(x))

∴ ¬∃x (D(x) ∧ M(x))

The conflict isn't considerable. Consider a class that includes a lone understudy named Frank. If M(Frank) = D(Frank) = T and A(Frank) = F, by then the hypotheses are overall evident and the end is counterfeit. Toward the day's end, Frank got a control, missed class, and didn't get an A.

Each understudy who missed class got a confinement.

Penelope is an understudy in the class.

Penelope got a confinement.

Penelope missed class.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got a confinement.

Each understudy who missed class got a confinement.

Penelope is an understudy in the class.

Penelope didn't miss class.

Penelope didn't get imprisonment.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got imprisonment.

Each understudy who missed class or got imprisonment didn't get an A.

Penelope is an understudy in the class.

Penelope got an A.

Penelope didn't get repression.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got a repression.

An (ax): x got an A.

H(x): x is on the regard roll

An (x): x got an A.

D(x): x got a repression.

∀x (H(x) → A(x)) a

¬∃x (D(x) ∧ A(x))

∴ ¬∃x (D(x) ∧ H(x))

Real.

1. ∀x (H(x) → A(x)) Hypothesis

2. c is a self-self-assured element Element definition

3. H(c) → A(c) Universal dispatch, 1, 2

4. ¬∃x (D(x) ∧ A(x)) Hypothesis

5. ∀x ¬(D(x) ∧ A(x)) De Morgan's law, 4

6. ¬(D(c) ∧ A(c)) Universal dispatch, 2, 5

7. ¬D(c) ∨ ¬A(c) De Morgan's law, 6

8. ¬A(c) ∨ ¬D(c) Commutative law, 7

9. ¬H(c) ∨ A(c) Conditional character, 3

10. A(c) ∨ ¬H(c) Commutative law, 9

11. ¬D(c) ∨ ¬H(c) Resolution, 8, 10

12. ¬(D(c) ∧ H(c)) De Morgan's law, 11

13. ∀x ¬(D(x) ∧ H(x)) Universal hypothesis, 2, 12

14. ¬∃x (D(x) ∧ H(x)) De Morgan's law, 13

4 0

3 years ago