Write a program that takes in a positive integer as input, and outputs a string of 1's and 0's representing the integer in binar
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Answer:
see explaination
Explanation:
We will consider the five Actors in the prearranged Online shopping software system:
Customer:
The scheme allows the customer to do below mention actions:
View goods and inquire about the niceties of products and their ease of use.
To create version to be able to purchase invention from the structure.
Browse crop through search category or shortest search alternative.
Place order for the necessary crop.
Make sum for the order(s) positioned.
Payment System:
Payment system allows client to pay using subsequent two methods:
Credit card.
PayPal.
Seller:
System allow seller to perform underneath actions:
Place the foodstuffs for selling under apposite product category.
Create account to happen to a member.
Administrator:
Following actions are perform by Admin:
Manage the goods posted in the system.
Modify existing manufactured goods categories or add novel categories for foodstuffs.
Site Manager:
System privileges Site director with the following role in the system:
View information on:
Orders Placed by customer
Products added by seller
Accounts created by users
check attachment
Answer:
<u>The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is </u><u>25.11 ms</u>
Explanation:
Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:
time taken 1=
= (56 x 8) bits / 64 x 10³ bits/s
= 7 x 10⁻³s
time taken 1= 7 ms
The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:
time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s
= 1.12 x 10⁻⁴ s
time taken 2= 112 μs
The propagation delay is 18 ms.
To calculate the total time elapsed, we need to add up all the time taken at each individual stage.
<u /><u> = Time taken 1 + Time taken 2 + Propagation Delay</u>
= 7 ms + 112 μs + 18 ms
= 0.025112 s
= 25.11 ms
Answer:
<h2>Function 1:</h2>#include <stdio.h> //for using input output functions
// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter
void PrintPopcornTime(int bagOunces){
if (bagOunces < 3){ //if value of bagOunces is less than 3
printf("Too small"); //displays Too small message in output
printf("\n"); } //prints a new line
//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10
else if (bagOunces > 10){
printf("Too large"); //displays the message: Too large in output
printf("\n"); //prints a new line }
/*the following else part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */
else {
/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */
//int result;
//result = bagOunces * 6;
//printf("%d",result);
printf("%d",bagOunces * 6); /multiplies value of bagOunces to 6
printf(" seconds");
// seconds is followed with the value of bagOunces * 6
printf("\n"); }} //prints a new line
int main(){ //start of main() function body
int userOunces; //declares integer variable userOunces
scanf("%d", &userOunces); //reads input value of userOunces
PrintPopcornTime(userOunces);
//calls PrintPopcornTime function passing the value in userOunces
return 0; }
Explanation:
<h2>Function 2: </h2>#include <stdio.h> //header file to use input output functions
// start of the function PrintShampooInstructions body having integer variable numCycles as parameter
void PrintShampooInstructions(int numCycles){
if(numCycles < 1){
//if conditions checks value of numCycles is less than 1 or not
printf("Too few."); //prints Too few in output if the above condition is true
printf("\n"); } //prints a new line
//else if part is executed when the if condition is false and else if checks //value of numCycles is greater than 4 or not
else if(numCycles > 4){
//prints Too many in output if the above condition is true
printf("Too many.");
printf("\n"); } //prints a new line
//else part is executed when the if and else if conditions are false
else{
//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done
for(int N = 1; N <= numCycles; N++){
printf("%d",N);
printf(": Lather and rinse. \n");}
printf("Done.");
printf("\n");} }
int main() //start of the main() function body
{ int userCycles; //declares integer variable userCycles
scanf("%d", &userCycles); //reads the input value into userCycles
PrintShampooInstructions(userCycles);
//calls PrintShampooInstructions function passing the value in userCycles
return 0;}
I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by
printf(": Lather and rinse. \n"); which is followed by printf("Done.");
So at first iteration:
printf("%d",N); prints 1 as the value of N is 1
printf(": Lather and rinse. \n"); prints : Lather and rinse and prints a new line \n.
As a whole this line is printed on the screen:
1: Lather and rinse.
Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.
Now at second iteration:
The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles which means 2=2. So the program control enters the body of loop.
printf("Done."); prints Done after the above two lines.
printf("%d",N); prints 2 as the value of N is 2
printf(": Lather and rinse. \n"); prints : Lather and rinse and prints a new line \n.
As a whole this line is printed on the screen:
2: Lather and rinse.
Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.
The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles which means 3>2. So the loop breaks.
Now the next statement is:
printf("Done."); which prints Done on the screen.
So as a whole the following output is displayed on the screen:
1: Lather and rinse.
2: Lather and rinse.
Done.
The programs along with their outputs are attached.
