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Salsk061 [2.6K]
2 years ago
10

The alternative to encapsulating security protocol (esp) is _________.

Computers and Technology
1 answer:
Lina20 [59]2 years ago
8 0
The alternative to Encapsulating Security Protocol (ESP) is Authentication Header (AH). Hope this helped!
You might be interested in
You are given an array x of string elements along with an int variable n that contains the number of elements in the array. You
Slav-nsk [51]

Answer:

The code to this question can be given as:

Code:

int i,j,count_previous=0,count_next=0; //define variable

for (j=0; j<n; j++) //loop for array

{

if (x[0]==x[j]) //check condition

{

count_previous++; //increment value by 1.

}

}

for (i=0; i<n; i++) //loop

{

for (j=0; j<n; j++)

{

if (x[i]==x[j]) //check condition

{

count_next++; //increment value by 1.

}

}

if (count_previous>count_next) //check condition

{

mode=x[i-1];

}

else

{

mode=x[i];

count_previous=count_next; //change value.

count_next= 0 ; //assign value.

}

}

Explanation:

In the question it is define that x is array of the string elements that is already define in the question so the code for perform operation in the array is given above. In the code firstly we define the variable that is i, j, count_previous, count_next. The variable i,j is used in the loop and variable count_previous and count_next we assign value 0 that is used for check the values of array. Then we define the for loop in this loop we use conditional statement in the if block we check that array of zero element is equal to array of j value then the count_previous is increase by 1. Then we use nested loop It is also known as loop in a loop. In this first loop is used for array and the second loop is used for check array element.In this we use the condition that if array x of i value is equal to array x of j then count_next will increment by 1.Then we use another condition that is if count_previous is greater then count_next then mode of x is decrement by 1. else mode equal to array and count_previous holds the value of count_next and count_next is equal to 0.

8 0
2 years ago
Olivia creates a personal budget. She enters her current savings account balance in cell D3. In cell A3, she calculates her inco
alexdok [17]

Answer is :

=D3+A3-B3

These Cell address contain following statement:

D3= Olivia saving account balance

A3= Olivia 3 month income

B3= Olivia 3 month expense

so, Expense are also deducted from income and income is added into saving balance.

In this situation we are use this formula:

=D3+A3-B3

5 0
1 year ago
What technique creates different hashes for the same password? ccna routing protocols final answers?
elena-s [515]
The answer is Salted Password Hashing.  The process is similar to hashing., but with a twist. A random value is introduced for each user. This salt value<span> is included with the password when the hash value is calculated and is stored with the user record. Including the salt value means that two users with the same password will have different password hashes.</span>
7 0
2 years ago
You're the sole IT employee at your company, and you don't know how many users or computers are in your organization. Uh oh! Wha
Neporo4naja [7]

Answer

Directory Services

<u></u>

<u>Definition</u>

It is a type of software that is used to unify and customize the resources available at the network.

<u>Explanation</u>

This software is used to gather the information and addresses of all computers connected on the network.

It is a type of information store, where all the information related to resources (computer) and services of network has been stored. It also gives administrator rights to the single point to manage whole network resources.

This is the reason, Option "D" is the best choice for me as sole IT employ in a company. I will prefer directory services software to gather the information of all computers of organization and manage users in company.  

4 0
2 years ago
Create a single list that contains the following collection of data in the order provided:
ivann1987 [24]

Answer:

A code was created to single list that contains the following collection of data in the order provided

Explanation:

Solution

#list that stores employee numbers

employeee_numbers=[1121,1122,1127,1132,1137,1138,1152,1157]

#list that stores employee names

employee_name=["Jackie Grainger","Jignesh Thrakkar","Dion Green","Jacob Gerber","Sarah Sanderson","Brandon Heck","David Toma","Charles King"]

#list that stores wages per hour employee

hourly_rate=[22.22,25.25,28.75,24.32,23.45,25.84,22.65,23.75]

#list that stores salary employee

company_raises=[]

max_value=0

#list used to store total_hourly_rate

total_hourly_rate=[]

underpaid_salaries=[]

#loop used to musltiply values in hourly_wages with 1.3

#and store in new list called total_hourly_rate

#len() is used to get length of list

for i in range(0,len(hourly_rate)):

#append value to the list total_hourly_rate

total_hourly_rate.append(hourly_rate[i]*1.3)

for i in range(0,len(total_hourly_rate)):

#finds the maximum value

if(total_hourly_rate[i]>max_value):

max_value=total_hourly_rate[i] #stores the maximum value

if(max_value>37.30):

print("\nSomeone's salary may be a budget concern.\n")

for i in range(0,len(total_hourly_rate)):

#check anyone's total_hourly_rate is between 28.15 and 30.65

if(total_hourly_rate[i]>=28.15 and total_hourly_rate[i]<=28.15):

underpaid_salaries.append(total_hourly_rate[i])

for i in range(0,len(hourly_rate)):

#check anyone's hourly_rate is between 22 and 24

if(hourly_rate[i]>22 and hourly_rate[i]<24):

#adding 5%

hourly_rate[i]+=((hourly_rate[i]/100)*5)

elif(hourly_rate[i]>24 and hourly_rate[i]<26):

#adding 4%

hourly_rate[i]+=((hourly_rate[i]/100)*4)

elif(hourly_rate[i]>26 and hourly_rate[i]<28):

#adding 3%

hourly_rate[i]+=((hourly_rate[i]/100)*3)

else:

#adding 2% for all other salaries

hourly_rate[i]+=((hourly_rate[i]/100)*2)

#adding all raised value to company_raises

company_raises.extend(hourly_rate)

#The comparison in single line

for i in range(0,len(hourly_rate)):

 For this exercise i used ternary

operator hourly_rate[i]= (hourly_rate[i]+((hourly_rate[i]/100)*5)) if (hourly_rate[i]>22 and hourly_rate[i]<24) else (hourly_rate[i]+((hourly_rate[i]/100)*4)) if (hourly_rate[i]>24 and hourly_rate[i]<26) else hourly_rate[i]+((hourly_rate[i]/100)*2)

8 0
2 years ago
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