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Dmitry [639]
3 years ago
12

Which type of denial of service attack exploits the existence of software flaws to disrupt a service?

Computers and Technology
1 answer:
hichkok12 [17]3 years ago
4 0
<span>The denial of service attack that exploits the existence of software flaws to disrupt a service is the teardrop attack. </span>The teardrop attack exploits flaws in the way older operating systems handled fragmented Internet Protocol (IP) packets.
<span>Fragmented packets are sent to a target machine but the target machine  cannot reassemble them due to a bug in TCP/IP fragmentation reassembly, the packets overlap one another. This results in crash in the target network device. </span>
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I used computers for 3 years now and i think its B
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2 years ago
Please help me on this coding problem :)
Vesnalui [34]

Answer:

a=4 , b=1

Explanation:

I'm not a computer science major at all but I think I can help you with this code.

Our program wants us to add 2 to a get new a value while also subtracting 1 from b value to obtain new b value. We we want to for for as long b is not 0 and a/b is nonnegative.

One round we get:

New a=0+2=2

New b=3-1=2

Let's see if we can go another round:

New a=2+2=4

New b=2-1=1

We can't go another round because b would be negative while a is positive which would make a/b negative. So our loop stops at this 2nd round.

a=4 , b=1

Other notes:

2nd choice makes no sense because a is always going to increase because of the addition on a and b was going to decrease because of the subtraction on it.

Third choice makes no sense because a/b doesn't even exist.

Fourth choice a/b is negative not nonnegative.

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2 years ago
Object Linking and Embedding (OLE) data integration allows you to
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Suppose a company A decides to set up a cloud to deliver Software as a Service to its clients through a remote location. Answer
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8 0
3 years ago
Assume the following variable definition appears in a program:
AnnZ [28]

Answer:

cout << setprecision(2)<< fixed << number;

Explanation:

The above statement returns 12.35 as output

Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.

The statement starts by setting precision to 2 using setprecision(2)

This is immediately followed by the fixed manipulator;

The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;

Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.

The fixed manipulator is then followed by the variable to be printed.

See code snippet below

<em>#include <iostream>  </em>

<em>#include <iomanip> </em>

<em>using namespace std;  </em>

<em>int main()  </em>

<em>{  </em>

<em> // Initializing the double value</em>

<em> double number = 12.3456;  </em>

<em> //Print  result</em>

<em> cout << setprecision(2)<< fixed << number;  </em>

<em> return 0;  </em>

<em>}  </em>

<em />

4 0
3 years ago
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