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:
int withinArray(int * intArray, int size, int * ptr) {
if(ptr == NULL) // if ptr == NULL return 0
return 0;
// if end of intArr is reached
if(size == 0)
return 0; // element not found
else
{
if(*(intArray+size-1) == *(ptr)) // check if (size-1)th element is equal to ptr
return 1; // return 1
return withinArray(intArray, size-1,ptr); // recursively call the function with size-1 elements
}
}
Explanation:
Above is the completion of the program.
Answer:
Explanation:
public class Temperature
{
double ftemp;
public int Constructor(double fahrenheit)
{
ftemp = fahrenheit;
return Convert.ToInt32(ftemp);
}
public void setFahrenheit(double fahrenheit)
{
ftemp = fahrenheit;
}
public void getFahrenheit()
{
ftemp = Constructor(ftemp);
}
public void getCelcius()
{
ftemp = (ftemp - 32) * 5 / 9;
}
public void getKelvin()
{
ftemp = (ftemp - 32) * 5 / 9 + 273.15;
}
}
