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Thepotemich [5.8K]
2 years ago
10

As the security administrator for your organization, you must be aware of all types of attacks that can occur and plan for them.

Which type of attack uses more than one computer to attack the victim?
Computers and Technology
1 answer:
Aliun [14]2 years ago
5 0

Answer:

The answer is "DDoS "

Explanation:

The distributed denial of service attack (DDoS) occurs whenever a directed program's wireless data or assets, generally one or even more application servers, were also swamped by various machines. This attack is always the consequence of many affected systems, that fill up the target network with traffic.

  • This attack is aimed to avoid legal customers of one's website from accessing it.  
  • In being effective in a DDoS attack, further demands need to be sent to the hacker than even the victim's server could deal with.  
  • One other way to successfully attack is to send fake requests from the attacker.
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Which section is optional in a resume?
suter [353]

<em>Career Objectives </em>are pointless in a resume, Things like experience, Contact details, education and skills, help the company decide whether to take you or not, in accordance with their needs. They don't need to know why you want the job as opposed to how you could help them.

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Sarah is a busy real estate agent with a growing clientele. She is looking to purchase a new computer and software so that she c
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2 years ago
What report would best help identify potential browser issues with website traffic?
Bezzdna [24]

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<span>a)      </span>The Active Users report

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<span>c)       </span>The Browser & OS report

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The answer is c) The browser & OS report


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6 0
3 years ago
Whers the main characteristic of a Peer-to-peer (P2P) network?
ioda

The answer to this would be B

5 0
3 years ago
Below is a list of 32-bit memory address references, given as word addresses.
Maru [420]

For a direct mapped cache the general rule is: first figure out the bits of the offset (the right-most bits of the address), then figure out the bits of the index (the next-to right-most address bits), and then the tag is everything left over (on the left side).

One way to think of a direct mapped cache is as a table with rows and columns. The index tells you what row to look at, then you compare the tag for that row, and if it matches, the offsettells you which column to use. (Note that the order you use the parts: index/tag/offset, is different than the order in which you figure out which bits are which: offset/index/tag.)

So in part (a) The block size is 1 word, so you need 0 offset bits (because <span><span><span>20</span>=1</span><span><span>20</span>=1</span></span>). You have 16 blocks, so you need 4 index bits to give 16 different indices (because <span><span><span>24</span>=16</span><span><span>24</span>=16</span></span>). That leaves you with the remaining 28 bits for the tag. You seem to have gotten this mostly right (except for the rows for "180" and "43" where you seem to have missed a few bits, and the row for "181" where you interchanged some bits when converting to binary, I think). You are correct that everything is a miss.

For part (b) The block size is 2 words, so you need 1 offset bit (because <span><span><span>21</span>=2</span><span><span>21</span>=2</span></span>). You have 8 blocks, so you need 3 index bits to give 8 different row indices (because <span><span><span>23</span>=8</span><span><span>23</span>=8</span></span>). That leaves you with the remaining 28 bits for the tag. Again you got it mostly right except for the rows for "180" and "43" and "181". (Which then will change some of the hits and misses.)

5 0
3 years ago
Read 2 more answers
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