Raymond should compress the files, and then at the very least ask what would be the appropriate time to email the attachment. It is important to practice good etiquette when using email. If you are sending an email that has attachments, always make sure to mention it in the body of the email. Raymond should also ZIP or compress his large attachments before sending them. This will make them easier to send.

4 0

3 years ago

Read 2 more answers

Write a program that reads in an integer value for n and then sums the integers from n to 2 * n if n is nonnegative, or from 2 *

devlian [24]

Answer:

Following are the program in c++

First code when using for loop

#include<iostream> // header file

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n

cin>>n;

if(n > 0) // check if n is nonnegative

{

for(i=n; i<=(2*n); i++)

{

sum1+= i;

}

else // check if n is negative

{

for(i=(2*n); i<=n; i++)

{

sum1+= i;

}

cout<<"The sum is:"<<sum1;

return 0;

Output

Enter the number: 1

The sum is:3

second code when using while loop

#include<iostream> // header file

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n

cin>>n;

if(n > 0) // check if n is nonnegative

{

int i=n;

while(i<=(2*n))

{

sum1+= i;

i++;

}

else // check if n is negative

{

int i=n;

while(i<=(2*n))

{

sum1+= i;

i++;

}

cout<<"The sum is:"<<sum1;

return 0;

}

Output

Enter the number: 1

The sum is:3

Explanation:

Here we making 2 program by using different loops but using same logic

Taking a user input in "n" variable.

if n is nonnegative then we iterate the loops n to 2 * n and storing the sum in "sum1" variable.

Otherwise iterate a loop 2 * n to n and storing the sum in "sum1" variable. Finally display sum1 variable .

4 0

3 years ago

Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th

ANEK [815]

Answer:

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112 ≈ 7/4

fractional value ( F ) = 0.112

And, numeric value of the quantity ( V ) = 7

The exponent bits will be represented as : 100----01.

while The fraction bits will be represented as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

= 1.11------12 = 2 - 2-n

also the fractional value ( F ) =

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E ) = 2k-1 – 2

While the significand value ( M ) = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112 ≈ 7/4

fractional value ( F ) = 0.112

And, numeric value of the quantity ( V ) = 7

The exponent bits will be represented as : 100----01.

while The fraction bits will be represented as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

= 1.11------12 = 2 - 2-n

also the fractional value ( F ) =

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E ) = 2k-1 – 2

While the significand value ( M ) = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

8 0

3 years ago