The answer is B!
Hope this helps.
Answer:
import java.util.Arrays;
import java.util.Scanner;
public class num1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter length of the array:");
int len = in.nextInt();
double [] temps = new double[len];
double avgTem;
int k =0;
double total = 0;
for( k=0; k<temps.length; k++){
System.out.println("Enter values for the array");
temps[k]=in.nextDouble();
}
System.out.println("The Arrays contains the following values");
System.out.println(Arrays.toString(temps));
// Computing the average of the values
for(k=0; k<temps.length; k++){
total = total+temps[k];
}
avgTem = total/(temps.length);
System.out.println("The average Temperature is: "+avgTem);
}
}
Explanation:
- Using Java programming language
- Import the Scanner class to receive user input
- Prompt User for the length of the Array, receive and store in a variable len;
- Declare a new double array of size len double [] temps = new double[len];
- Using a for loop, continually prompt user to enter values into the array
- Display the values of the array using Java's Arrays.toString method
- Use another for loop to add up all the elements in the arraay and store in the variable called total
- Outside the second for loop calculate the average avgTem = total/(temps.length);
- Display the average temp.
Answer:
The answer is 500 kbps
Explanation:
Consider the given data in the question.
R1 = 500 kbps
R2=2 Mbps
R3 = 1 Mbps
Now as it is mentioned that there is no other traffic in the network.
Thus,
throughput of the file = min {R1,R2,R3}
throughput of the file = min {500 kbps, 2 Mbps, 1 Mbps}
T/P of the file = 500 kbps
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Explanation:
