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2 years ago

What elements of SANS 20 you could leverage to reduce the TCP/IP vulnerabilities of your workstation

Computers and Technology

1 answer:

Vlada [557]2 years ago

5 0

Answer:

- limitation and control of network ports, protocols and services.

- Continuous vulnerability assessment and remediation.

Explanation:

The TCP or transmission control protocol is a layer 4 protocol (transport), that reliably transports packets in sequential segments to an application in the destination computer, using the ip address and the port number of the application.

The ICMP sent during the TCP/ip activities, can render the network vulnerable to attacks. limitation and control of the network ports, protocols and services and continuous assessment would mitigate the vulnerability of the TCP/ip model.

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On a webpage, a _____ provides supplemental material such as social networking feeds and ads. Group of answer choices footer sid

finlep [7]

Answer:

sidebar

Explanation:

The part of the webpage that is being described in this question is known as a sidebar. These are the far left/right sides of the webpage and are usually used for networking feeds, ads, related information, major points of the main page, biographical info, as well as persuasive information. This section of the webpage serves two main purposes to provide the viewer with advertisements and to encourage the reader to read the more detailed main article.

8 0

2 years ago

Write a copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *thi

Olin [163]

The following cose will be used to copy assignment operator for CarCounter

<u>Explanation:</u>

Complete Program:

#include <iostream>

using namespace std;

class CarCounter

{

public:

CarCounter();

CarCounter& operator=(const CarCounter& objToCopy);

void SetCarCount(const int setVal)

{

carCount = setVal;

}

int GetCarCount() const

{

return carCount;

}

private:

int carCount;

};

CarCounter::CarCounter()

{

carCount = 0;

return;

}

// FIXME write copy assignment operator

/* Your solution goes here */

CarCounter& CarCounter::operator=(const CarCounter& objToCopy)

{

if(this != &objToCopy)

carCount = objToCopy.carCount;

return *this;

}

int main()

{

CarCounter frontParkingLot;

CarCounter backParkingLot;

frontParkingLot.SetCarCount(12);

backParkingLot = frontParkingLot;

cout << "Cars counted: " << backParkingLot.GetCarCount();

cout << endl << endl;

system("pause");

return 0;

}

5 0

2 years ago

When creating a table in the relational database design from an entity in the extended e-r model, the attributes of the entity b

larisa86 [58]

No. The attributes of the table correspond to columns within the table. Each unique set of attribute values taken together correspond to table rows. Sometimes referred to as "records".

8 0

2 years ago

Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:

horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed. So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

= 0.00833 * 100

= 0.833 ms

So there is 1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d) Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0

2 years ago

The elements in a string type array will be initialized to ____.?

ivanzaharov [21]

The elements in a string type array will be initialized to "Null".

7 0

2 years ago