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qwelly [4]
2 years ago
5

In computing, a(n) _____ is an attack on an information system that takes advantage of a particular system vulnerability. Select

one: a. exit door b. glitch c. bot d. exploit
Computers and Technology
1 answer:
tensa zangetsu [6.8K]2 years ago
4 0

Answer: d) Exploit

Explanation: Exploit is a type computer attack that successful when the computer system of an user is vulnerable and attacker can do the exploitation. This happens due to the weakness of the system, applications software, network etc.

Other given option are incorrect because exit door,glitch and bad are not any type of attack in the computer field that causes harm to the system.Thus the correct option is option(d).

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In public-key encryption, the two keys–one for coding and one for decoding–are known as ________.
Feliz [49]
Public key and private key  - In public key encryption, a pair of keys is used (public key and private key). The public key can be made available publicly, while the private key is only known  by the owner. The public key is used to encrypt the message, while the private key is used to decrypt the message.
5 0
2 years ago
Suppose your company has decided that it needs to make certain busy servers faster. Processes in the workload spend 60% of their
vekshin1

Answer:

CPU need 50% much faster

disk need 100% much faster

Explanation:

given data

workload spend time CPU  = 60%

workload spend time I/O = 40%

achieve overall system speedup = 25%

to find out

How much faster does CPU need and How much faster does the disk need

solution

we apply here Amdahl’s law for the overall speed of a computer that is express as

S = \frac{1}{(1-f)+ \frac{f}{k} }      .............................1

here f is fraction of work i.e 0.6 and S is overall speed  i.e 100% + 25% = 125 % and k is speed up of component

so put all value in equation 1 we get

S = \frac{1}{(1-f)+ \frac{f}{k} }  

1.25 = \frac{1}{(1-0.6)+ \frac{0.6}{k} }  

solve we get

k = 1.5

so we can say  CPU need 50% much faster

and

when f = 0.4 and S = 125 %

put the value in equation 1

S = \frac{1}{(1-f)+ \frac{f}{k} }  

1.25 = \frac{1}{(1-0.4)+ \frac{0.4}{k} }  

solve we get

k = 2

so here disk need 100% much faster

7 0
2 years ago
The microprogram counter (MPC) contains the address of the next microcode statement for the Mic1 emulator to execute. The MPC va
Ray Of Light [21]

Answer:

MAR bit

Explanation:

The MPC computers area set of software and hardware standards that was developed by the consortium of computer firms which is led by Microsoft. It contains the address of a next microcode for the Mic1 emulator foe execution.

The part of the executing microcode instruction that determines the value which is placed in the MOC is the MAR bit. The MAR is the memory address register in the CPU which stores the memory address or such addresses to which some data will be sent and also stored.

6 0
1 year ago
Adrian has decided to create a website to catalog all the books he has in his personal library. He wants to store this informati
Lostsunrise [7]
I think it is XML. Hope that helps.
8 0
2 years ago
Read 2 more answers
Write two methods: encrypt and decrypt. encrypt should #take as input a string, and return an encrypted version #of it according
Harman [31]

Answer:

The code is given below with appropriate comments

Explanation:

CIPHER = (("D", "A", "V", "I", "O"),

         ("Y", "N", "E", "R", "B"),

         ("C", "F", "G", "H", "K"),

         ("L", "M", "P", "Q", "S"),

         ("T", "U", "W", "X", "Z"))

# Add your code here!

def encrypt(plaintext):

   theList = []

   for char in plaintext:

       if char.isalpha():

           char = char.upper()

           if char == "J":

               char = "I"

           theList.append(char)

   if len(theList) % 2 == 1:

       theList.append("X")

   for i in range(0, len(theList), 2):

       if theList[i] == theList[i + 1]:

           theList[i + 1] = "X"

       findex = [-1, -1]

       sindex = [-1, -1]

       for j in range(len(CIPHER)):

           for k in range(len(CIPHER)):

               if theList[i] == CIPHER[j][k]:

                   findex = [j, k]

               if theList[i + 1] == CIPHER[j][k]:

                   sindex = [j, k]

       # same row

       if (findex[0] == sindex[0]):

           findex[1] += 1

           sindex[1] += 1

           if findex[1] == 5:

               findex[1] = 0

           if sindex[1] == 5:

               sindex[1] = 0

           theList[i] = CIPHER[findex[0]][findex[1]]

           theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

       # same column

       elif (findex[1] == sindex[1]):

           findex[0] += 1

           sindex[0] += 1

           if findex[0] == 5:

               findex[0] = 0

           if sindex[0] == 5:

               sindex[0] = 0

           theList[i] = CIPHER[findex[0]][findex[1]]

           theList[i + 1] = CIPHER[sindex[0]][sindex[1]]

       else:

           theList[i] = CIPHER[findex[0]][sindex[1]]

           theList[i + 1] = CIPHER[sindex[0]][findex[1]]

   return "".join(theList)

def decrypt(ciphertext):

   theString = ""

   findex = [-1, -1]

   sindex = [-1, -1]

   for i in range(0, len(ciphertext), 2):

       for j in range(len(CIPHER)):

           for k in range(len(CIPHER)):

               if ciphertext[i] == CIPHER[j][k]:

                   findex = [j, k]

               if ciphertext[i + 1] == CIPHER[j][k]:

                   sindex = [j, k]

       if (findex[0] == sindex[0]):

           findex[1] -= 1

           sindex[1] -= 1

           if findex[1] == -1:

               findex[1] = 4

           if sindex[1] == -1:

               sindex[1] = 4

           theString += CIPHER[findex[0]][findex[1]]

           theString += CIPHER[sindex[0]][sindex[1]]

       # same column

       elif (findex[1] == sindex[1]):

           findex[0] -= 1

           sindex[0] -= 1

           if findex[0] == -1:

               findex[0] = 4

           if sindex[0] == -1:

               sindex[0] = 4

           theString += CIPHER[findex[0]][findex[1]]

           theString += CIPHER[sindex[0]][sindex[1]]

       else:

           theString += CIPHER[findex[0]][sindex[1]]

           theString += CIPHER[sindex[0]][findex[1]]

   return theString

# Below are some lines of code that will test your function.

# You can change the value of the variable(s) to test your

# function with different inputs.

#

# If your function works correctly, this will originally

# print: QLGRQTVZIBTYQZ, then PSHELXOWORLDSX

print(encrypt("PS. Hello, worlds"))

print(decrypt("QLGRQTVZIBTYQZ"))

5 0
2 years ago
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