Answer:
B) Punctuation
Explanation:
Punctuation are greatly used in different programming languages for different purposes. In the Java, C++ and C# Programming languages for example, two popular punctuation marks are the comma (,) and semi-colon (;). The comma is used for separating elements of a list and arrays, while the semi-colon indicates end of an executable statement or line of code. Other popular punctuation marks used in programming are periods (dot), question marks, parenthesis angle brackets and braces each implementing a specific grammatical syntax in the language.
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Answer is salvation
Answer:
// program in C++.
// headres
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// array
int temperatures[7];
// count variable
int count=0;
cout<<"Enter the temperature of all days:";
for(int a=0;a<7;a++)
{
// read temperature of 7 days
cin>>temperatures[a];
// find temperature is extreme or not
if(temperatures[a]<-10||temperatures[a]>25)
// count
count++;
}
// print count of extreme temperature
cout<<"number of days of extreme temperature:"<<count<<endl;
return 0;
}
Explanation:
Create an array of size 7 to store the temperature of all days of week.Read the temperature of each day.If the temperature is less than -10 or greater than 25 then increment the count.This will count the number of days of extreme temperature.Print the count.
Output:
Enter the temperature of all days:-20 12 18 30 32 -15 15
number of days of extreme temperature:4
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
I don’t get this but you just put random number
