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Anna007 [38]
2 years ago
7

Users report that the network access is slow. After questioning the employees, the network administrator learned that one employ

ee downloaded a third-party scanning program for the printer. What type of malware might be introduced that causes slow performance of the network?
Computers and Technology
1 answer:
grigory [225]2 years ago
3 0

Keeping the fact in mind that network access has been found to be slow and after questioning the employees, the network administrator learns that an employee downloaded a third-party application which turns out to be a scanning program for the printer.

<u>Explanation:</u>

The employee must have been unaware of the fact that the application was accompanied by a worm or a computer worm.

A computer worm is a type of malware that has a property to replicate itself and after infecting a particular computer increases its range and gets in the other computers working on the same network which is the main reason for reduced processing speed.

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Calvin is an aspiring graphic designer. He wants to achieve Adobe Certified Expert certification in software that will be helpfu
bazaltina [42]

Answer:

a

Explanation:

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2 years ago
If byte stuffing is used to transmit Data, what is the byte sequence of the frame (including framing characters)? Format answer
Lerok [7]

Answer:

Correct Answers: 01h 79h 1Bh 78h 78h 1Bh 7Ah 04

Explanation:

Solution is attached below

4 0
2 years ago
This question involves the creation of user names for an online system. A user name is created based on a user’s first and last
Evgen [1.6K]

Answer:

See explaination

Explanation:

import java.util.*;

class UserName{

ArrayList<String> possibleNames;

UserName(String firstName, String lastName){

if(this.isValidName(firstName) && this.isValidName(lastName)){

possibleNames = new ArrayList<String>();

for(int i=1;i<firstName.length()+1;i++){

possibleNames.add(lastName+firstName.substring(0,i));

}

}else{

System.out.println("firstName and lastName must contain letters only.");

}

}

public boolean isUsed(String name, String[] arr){

for(int i=0;i<arr.length;i++){

if(name.equals(arr[i]))

return true;

}

return false;

}

public void setAvailableUserNames(String[] usedNames){

String[] names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

for(int i=0;i<usedNames.length;i++){

if(isUsed(usedNames[i],names)){

int index = this.possibleNames.indexOf(usedNames[i]);

this.possibleNames.remove(index);

names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

}

}

}

public boolean isValidName(String str){

if(str.length()==0) return false;

for(int i=0;i<str.length();i++){

if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

return false;

}

return true;

}

public static void main(String[] args) {

UserName person1 = new UserName("john","smith");

System.out.println(person1.possibleNames);

String[] used = {"harta","hartm","harty"};

UserName person2 = new UserName("mary","hart");

System.out.println("possibleNames before removing: "+person2.possibleNames);

person2.setAvailableUserNames(used);

System.out.println("possibleNames after removing: "+person2.possibleNames);

}

}

8 0
2 years ago
A 1.17 g sample of an alkane hydrocarbon gas occupies a volume of 674 mL at 28°C and 741 mmHg. Alkanes are known to have the gen
Nina [5.8K]

Answer:

C3H8

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of alkane = 1.17g

Volume (V) = 674 mL

Temperature (T) = 28°C

Pressure (P) = 741 mmHg.

Gas constant (R) = 0.08206 atm.L/Kmol

Step 2:

Conversion to appropriate unit.

For Volume:

1000mL = 1L

Therefore, 674mL = 674/1000 = 0.674L

For Temperature:

Temperature (Kelvin) = Temperature (celsius) + 273

Temperature (celsius) = 28°C

Temperature (Kelvin) = 28°C + 273 = 301K

For Pressure:

760mmHg = 1atm

Therefore, 741 mmHg = 741/760 = 0.975atm

Step 3:

Determination of the number of mole of the alkane..

The number of mole of the alkane can be obtained by using the ideal gas equation. This is illustrated below:

Volume (V) = 0.674L

Temperature (T) = 301K

Pressure (P) = 0.975atm

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

PV = nRT

n = PV /RT

n = (0.975 x 0.674)/(0.08206x301)

n = 0.0266 mole

Step 4:

Determination of the molar mass of the alkane.

Mass of alkane = 1.17g

Number of mole of the alkane = 0.0266mole

Molar Mass of the alkane =?

Number of mole = Mass/Molar Mass

Molar Mass = Mass/number of mole

Molar Mass of the alkane = 1.17/0.0266 = 44g/mol

Step 5:

Determination of the molecular formula of the alkane.

This is illustrated below:

The general formula for the alkane is CnH2n+2

To obtain the molecular formula for the alkane we shall assume n = 1, 2, 3 or more till we arrive at molar Mass of 44.

When n = 1

CnH2n+2 = CH4 = 12 + (4x1) = 16g/mol

When n = 2

CnH2n+2 = C2H6 = (12x2) + (6x1) = 30g/mol

When n = 3

CnH2n+2 = C3H8 = (12x3) + (8x1) = 44g/mol

We can see that when n is 3, the molar mass is 44g/mol.

Therefore, the molecular formula for the alkane is C3H8.

7 0
2 years ago
Select the correct text in the passage.
aliina [53]

Answer: When was admitted, hospital authorities recorded his medical history. Then, placed in an ICU where his vital signs were constantly monitored.

Explanation: With the help of computers, medical histories are often kept in the computer for future reference. Machines are connected to computers to record vital signs.

4 0
2 years ago
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