Answer:
No need to be overly-sensitive, brainly deleted that question
If no one joins then its not a problem
Here you go,
import java.util.Scanner;
import java.util.Random;
public class OrderCalculator{
public static void main(String[] args){
float x, y, z, semi, area;
Scanner in = new Scanner(System.in);
System.out.print("Enter the 3 sides: ");
x = in.nextFloat();
y = in.nextFloat();
z = in.nextFloat();
semi = (float) ((x + y + z) / 2.0);
area = (float) Math.sqrt(semi * (semi - x) * (semi - y) * (semi - z));
System.out.printf("The area is: %.3f\n", area);
}
}
Answer:
1) 402.7 grams. This estimate is called the sample mean.
2) (399.11, 406.29)
3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.
I am 99% sure that the value lies between 399.11 grams and 406.29 grams.
Explanation:
sample size (n) = 40, the mean weight (x)= 402.7 grams and the standard deviation (σ)=8.8 grams
1) The point estimated mean weight of the population is 402.7 grams. This estimate is called the sample mean.
2) c = 99% = 0.99
α = 1 - 0.99 = 0.01
.
The z score of 0.005 corresponds with the z score of 0.495 (0.5 - 0.005).
.
The margin of error (e) = 
The confidence interval = x ± e = 402.7 ± 3.59 = (399.11, 406.29)
3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.
I am 99% sure that the value lies between 399.11 grams and 406.29 grams.
Answer:
things is the correct answer
Answer:
.serial port
.VGA port
.Modem port
explain A port is a channel through which a device communicate with the operating system.