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serg [7]
2 years ago
8

Five routers are to be connected in a point-to-point subnet. between each pair of routers, the designers may put a high-speed li

ne, a medium-speed line, a low-speed line, or no line. if it takes 100 ms of computer time to generate and inspect each topology, how long will it take to inspect all of them?
Computers and Technology
1 answer:
Natalija [7]2 years ago
6 0

Let assume are lettered A to E in that order. Thus, there will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of these potential lines has 4 possibilities. Therefore, the total number of topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect each and one of them.</span>











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We have an internal webserver, used only for testing purposes, at IP address 5.6.7.8 on our internal corporate network. The pack
lukranit [14]

Answer:

Check the explanation

Explanation:

A packet filter firewall is used as a check point between internal corporate network to the outside internet. It blocks all the inbound traffic from the outside hosts trying to initiate a direct TCP connection to the internal corporate webserver. The network design with firewall is shown in the attached image below:

The figures in the attached image below shows an internal corporate network is protected with a packet filter firewall to minimize the inbound traffic from the external network or an internet. Therefore, the packet filter is used as a check point between the network.

The packet filter blocks all attempts by the outside hosts in order to initiate a direct TCP connection to the internal webserver of the internal corporate network.

Going by the second part of the attached image below can can therefore conclude that:

• Rule 1 specifies that, deny any packet with the destination address 5.6.7.8 if the STN flag of TCP header is set.

• Rule 2 specifies that, allow the inbound email traffic from the external source.

• Rule 3 specifies, allows the Outbound TCP traffic from the internal corporate network.

• Rule 4 specifies, allows outbound Email traffic from the internal corporate network to the external network.

• Rule 5 specifies, block any traffic from any source to the any destination.

3 0
2 years ago
Which would be the most efficient way to store files on your computer?
Tamiku [17]

Answer:

4

Explanation: All the other ones are a rushed way of storing files

5 0
2 years ago
Read 2 more answers
Assume that machines A and B are on the same network 10.3.2.0/24. Machine A sends out spoofed packets, and Machine B tries to sn
Zanzabum

Answer:

Explanation:

Assuming the spoofed 1.2.3.4, it will be sent out ARP to a router that is alive and get it MAC. Then the spoofed packet will be able to be sent out. If A spoofed 10.0.2.30, because it is a local address, it will send an ARP to the machine for been able to get the MAC. But, the IP address 10.0.2.30 is not original making it to be fake, so it will not replay to A, for this reason the spoofed packet cannot be sent out.

7 0
2 years ago
#Write a function called fancy_find. fancy_find should have #two parameters: search_within and search_for. # #fancy_find should
Inessa05 [86]

Answer:

Here is the Python program:

def fancy_find(search_within , search_for):  # function definition of fancy_find function that takes two parameters

   index = 0  #to store the index of search_within where the search_for string is found

   if search_for in search_within:  #checks if the string search_for is present in string search_within

       sf = search_for[0]  #points to the first character of the search_for

       for sw in search_within:  #iterates through search_within

           if sw == sf:  #if the first character of search_for is equal to the character at sw index of search_within

               if search_within[index:index+len(search_for)] == search_for:  #checks if the value of search_for is found in search_within

                   print(search_for,"found at index",index,"!")  #if above condition is true prints the message "[search_for] found at index [index]!", with [search_for] and [index] replaced by the value of search_for and the index at which it is found

                   return ""  

           index += 1  #increments value of index at each iteration

   print(search_for,"is not found within", search_within)  #if search_for is not found within search_within, prints message "[search_for] was not found within [search_within]!" with the values of search_for and search_within.

   return ""    

#following two statements are used to test the working of above function

print(fancy_find("ABCDEF", "DEF"))  #calls fancy_find() passing "ABCDEF" as search_within and "DEF" as search_for

print(fancy_find("ABCDEF", "GHI")) #calls fancy_find() passing "ABCDEF" as search_within and "GHI" as search_for

Explanation:

The program is well explained in the comments. I will explain the working of the function with the help of an example:

Suppose

search_within = "ABCDEF"

search_for = "DEF"

We have to find if search_for i.e. DEF is present in search_within i.e. ABCDEF

if search_for in search_within statement checks using in operator that if DEF is included in ABCDEF. Here this condition evaluates to true so the next statement sf = search_for[0]  executes which sets the first element of search_for i.e. D to sf. So sf = 'D'

for sw in search_within this statement has a for loop that iterates through ABCDEF and works as following:

At first iteration:

sw contains the first character of search_within  i.e. A

if sw == sf: condition checks if the first character of the search_for i.e. D is equal to sw i.e. A. Its not true so the program control moves to this statement:

index += 1 This increases the value of index by 1. index was initialized to 0 so now it becomes 1. Hence index=1

At second iteration:

sw contains the second character of search_within  i.e. B

if sw == sf: condition checks if the first character of the search_for i.e. D is equal to sw i.e. B Its not true so the program control moves to this statement:

index += 1 This increases the value of index by 1. index was initialized to 0 so now it becomes  2. Hence index=2

At third iteration:

sw contains the third character of search_within  i.e. C

if sw == sf: condition checks if the first character of the search_for i.e. D is equal to sw i.e. C Its not true so the program control moves to this statement:

index += 1 This increases the value of index by 1. index was initialized to 0 so now it becomes  3. Hence index=3

At fourth iteration:

sw contains the third character of search_within  i.e. D

if sw == sf: condition checks if the first character of the search_for i.e. D is equal to sw i.e. D. Its true so so the program control moves to this statement:

  if search_within[index:index+len(search_for)] == search_for:

current value of index=3

len(search_for) returns the length of DEF i.e. 3

So the if condition checks for the search_for in search_within. The statement becomes:

if search_within[3:3+3] == search_for:

search_within[3:3+3]  means from 3rd index position of search_within to 6-th index position of the search_within. This means from 4th element of search_within i.e. D to the last. Hence search_within[3:3+3] is equal to DEF.

search_for = DEF so

if search_within[3:3+3] == search_for: checks if

search_within[3:3+3] = DEF is equals to search_for = DEF

Yes it is true so

print(search_for,"found at index",index,"!") statement is executef which prints the following message:

DEF found at index 3!

This output is because search_for = "DEF" and index=3

5 0
2 years ago
If the computer has an encrypted drive, a ____ acquisition is done if the password or passphrase is available. a. passive b. sta
Aleonysh [2.5K]

Answer: C. Live

Explanation:

A live acquisition is where data is retrieved from a digital device directly via its normal interface such as activating a computer and initiating executables. Doing so has a certain level of risk because data is highly likely to be modified by the operating system. The process becomes significantly more common likely with less available disk space to the point of completely impractical to image.

7 0
1 year ago
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