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3 years ago

How to calculate 3 X (50 + 40) ÷ 5 on excel 2016

Computers and Technology

2 answers:

kozerog [31]3 years ago

8 0

Answer:

I think this might help, don't know much of this

Explanation:

How do you calculate 3.5 increase in Excel?

How To Increase a Number By a Percentage. If want to calculate a percentage increase in Excel (i.e. increase a number by a specified percentage), this can be done by simply multiply the number by 1 + the percentage increase. - which gives the result 60.

Juliette [100K]3 years ago

7 0

Explanation:

I am not sure what your problem is.

you don't know how to enter a formula into an Excel cell ? or you don't know how to combine the content of cells into one result in another cell ? or ... ?

but just answering literally this one-line-problem-definition

you click into an empty cell and type

=3*(50+40)/5

and press return.

and the cell will show you the result. 54

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A Windows user called the help desk to request that her local user account password be reset on her computer. The help desk tech

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Answer:

C. The user had previously encrypted her data files and folders using EFS.

Explanation:

The most likely reason for the lost data files is that the user had previously encrypted her data files and folders using EFS.

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3 years ago

Implement the RC4 stream cipher in C++. User should be able to enter any key that is 5 bytes to 32 bytes long. Be sure to discar

gladu [14]

Answer:

Explanation:

#include <iostream>

#include <string>

#include<vector>

using namespace std;

vector<int> permute(vector<int>, vector<int>);

string encrypt(vector<int>s1 , vector<int> t1, string p);

string decrypt(vector<int>s1, vector<int> t1, string p);

int main() {

string plaintext = "cryptology";

string plaintext2 = "RC4";

vector<int> S(256);

vector<int> T(256);

int key[] = { 1,2,3,6 };

int key2[] = { 5,7,8,9 };

int tmp = 0;

for (int i = 0; i < 256;i++) {

S[i] = i;

T[i] = key[( i % (sizeof(key)/sizeof(*key)) )];

}

S = permute(S, T);

for (int i = 0; i < 256 ;i++) {

cout << S[i] << " ";

if ((i + 1) % 16 == 0)

cout << endl;

}

cout << endl;

string p = encrypt(S, T, plaintext);

cout << "Message: " << plaintext << endl;

cout << "Encrypted Message: " << " " << p << endl;

cout << "Decrypted Message: " << decrypt(S, T, p) << endl << endl;

tmp = 0;

for (int i = 0; i < 256;i++) {

S[i] = i;

T[i] = key2[(i % (sizeof(key) / sizeof(*key)))];

}

S = permute(S, T);

for (int i = 0; i < 256;i++) {

cout << S[i] << " ";

if ((i + 1) % 16 == 0)

cout << endl;

}

cout << endl;

p = encrypt(S, T, plaintext2);

cout << "Message: " << plaintext2 << endl;

cout << "Encrypted Msg: " << p << endl;

cout << "Decrypted Msg: "<<decrypt(S, T, p) << endl << endl;

return 0;

}

string decrypt(vector<int>s1, vector<int> t1, string p) {

int i = 0;

int j = 0;

int tmp = 0;

int k = 0;

int b;

int c;

int * plain = new int[p.length()];

string plainT;

for (int r = 0; r < p.length(); r++) {

i = (i + 1) % 256;

j = (j + s1[i]) % 256;

b = s1[i];

s1[i] = s1[j];

s1[j] = b;

tmp = (s1[i] + s1[j]) % 256;

k = s1[tmp];

c = ((int)p[r] ^ k);

plain[r] = c;

plainT += (char)plain[r];

}

return plainT;

}

string encrypt(vector<int>s1, vector<int> t1, string p) {

int i = 0;

int j = 0;

int tmp = 0;

int k = 0;

int b;

int c;

int * cipher = new int [p.length()];

string cipherT;

cout << "Keys Generated for plaintext: ";

for (int r = 0; r < p.length(); r++) {

i = (i + 1) % 256;

j = (j + s1[i]) % 256;

b = s1[i];

s1[i] = s1[j];

s1[j] = b;

tmp = (s1[i] + s1[j]) % 256;

k = s1[tmp];

cout << k << " ";

c = ((int)p[r] ^ k);

cipher[r] = c;

cipherT += (char)cipher[r];

}

cout << endl;

return cipherT;

}

vector<int> permute(vector<int> s1, vector<int> t1) {

int j = 0;

int tmp;

for (int i = 0; i< 256; i++) {

j = (j + s1[i] + t1[i]) % 256;

tmp = s1[i];

s1[i] = s1[j];

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4 years ago

Which of the following devices electronically sorts mail by ZIP code?

irina [24]

The answer is OCR or optical character reader.
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3 years ago

4-Translate the following C program to MIPS assembly program (Please explain each instruction in your code by a comment and subm

earnstyle [38]

Answer:

.data

str1: .asciiz "Hello"

str2: .asciiz "mars"

msg: .asciiz "String after the concatenate: "

.text

main:

#load the address str1 to $a0

la $a0,str1

#initialize $t0 with 0

li $t0,0

#loop to find the length of str1

#$t0 stores the length of str1

loop1:

#load byte of $a0 to $t1

lb $t1,0($a0)

#branch for equal.If $t1 equal to 0,jump to label stop1

beq $t1,0,stop1

addi $t0,$t0,1 #increase the count

addi $a0,$a0,1 #increase the address

#jump to label loop1

j loop1

stop1:

#load the address of str2 to $a1

la $a1,str2

#loop2 concatenate the each element in str2 to str1

loop2:

lb $t2,0($a1) #load character in str2 to $t2

sb $t2,0($a0) #store value in $t2 to str1

beq $t2,0,stop2

addi $a1,$a1,1 #increase the address of str2

addi $a0,$a0,1 #increase the address of str1

j loop2

stop2:

#syscall for print string is $v0 = 4

#syscall to termiante the program is $v0 = 10

#print the message

li $v0, 4

la $a0, msg

syscall

#print the concatenated string

li $v0,4

la $a0,str1

syscall

#termiante the program

li $v0, 10

syscall

Explanation:

3 0

3 years ago

Isabel has an interesting way of summing up the values in a sequence A ofn integers, where n is a power of two. She creates a ne

olganol [36]

Answer:

Running time of algorithm is O(n).

Explanation:

n is power of 2

n =2,4,8,16,32,...................................

A is an array having n elements

B is an array of size 0 to (n/2)-1

if n=4 B then (4/2)-1 =1 So B has size 2

for(i=0;i<=(n/2)-1;++)

{

B[i]=A[2i]+A[2i+1];

}

This for loop will run n/2 times so complexity in terms of Big Oh is O(n/2) =O(n)

Running time of algorithm is O(n).

6 0

3 years ago