Answer:
Explanation:
#include <stdio.h>
int main(void)
{
int num, rem;
printf("Enter a number: ");
scanf("%d", &num);
printf("Roman numerals: ");
while(num != 0)
{
if (num >= 1000) // 1000 - m
{
printf("m");
num -= 1000;
}
else if (num >= 900) // 900 - cm
{
printf("cm");
num -= 900;
}
else if (num >= 500) // 500 - d
{
printf("d");
num -= 500;
}
else if (num >= 400) // 400 - cd
{
printf("cd");
num -= 400;
}
else if (num >= 100) // 100 - c
{
printf("c");
num -= 100;
}
else if (num >= 90) // 90 - xc
{
printf("xc");
num -= 90;
}
else if (num >= 50) // 50 - l
{
printf("l");
num -= 50;
}
else if (num >= 40) // 40 - xl
{
printf("xl");
num -= 40;
}
else if (num >= 10) // 10 - x
{
printf("x");
num -= 10;
}
else if (num >= 9) // 9 - ix
{
printf("ix");
num -= 9;
}
else if (num >= 5) // 5 - v
{
printf("v");
num -= 5;
}
else if (num >= 4) // 4 - iv
{
printf("iv");
num -= 4;
}
else if (num >= 1) // 1 - i
{
printf("i");
num -= 1;
}
}
return 0;
}
Answer:
The answer is below
Explanation:
Given that:
Frame transmission time (X) = 40 ms
Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request
a) Probability that there is success on the first attempt = 
but k = 0, therefore Probability that there is success on the first attempt = 
b) probability of exactly k collisions and then a success = P(collisions in k attempts) × P(success in k+1 attempt)
P(collisions in k attempts) = [1-Probability that there is success on the first attempt]^k = ![[1-e^{-G}]^k=[1-0.135]^k=0.865^k](https://tex.z-dn.net/?f=%5B1-e%5E%7B-G%7D%5D%5Ek%3D%5B1-0.135%5D%5Ek%3D0.865%5Ek)
P(success in k+1 attempt) =
Probability of exactly k collisions and then a success = 
c) Expected number of transmission attempts needed = probability of success in k transmission = 
