Answer:
the equipment are needed for wind power electrical generation are:-gabage,
water and
D. M water
Answer:
Java program given below
Explanation:
import java.util.Scanner;
public class LeapYearCheck {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
while(true)
{
System.out.print("Enter a year (0 to stop): ");
int year=in.nextInt();
// System.out.println();
if(year==0)
{
System.out.println("Bye.");
break;
}
boolean leap = false;
if(year % 4 == 0)
{
if( year % 100 == 0)
{
if ( year % 400 == 0)
leap = true;
else
leap = false;
}
else
leap = true;
}
else
leap = false;
if(leap)
System.out.println(year + " is a leap year.");
else
System.out.println(year + " is not a leap year.");
}
}
}
The true statement about the dot plot is 1 has 4 and 0 dots.
Explanation:
- after creating a Bar chart 4 is the right answer.
- The highest among st all the other plots is 1 but 4 shows 3.
- Taking an average from all the data points 3 comes to the right answer.
- Median the central Mid-point is 3.
- Mode also comes to 3.
- It is skewed on the right due to the 1st one.
- Skewed shows the data point either increasing or in decreasing.
- There a to Bi- histograms which has two ups and downs.
- The true statement has to be as Mean=Median=Mode is 3.
Answer:
Both are incorrect.
Explanation:
PCV valve opening is dependent on amount of manifold vacuum value. The opening cannot decrease due to part throttle operation. The PCV system is a tampered valve whose opening depend upon intake manifold vacuum. PCV valve is supported by a spring and is initially in a closed position.
Answer:
//Convert any decimal number to binary number
//Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include <iostream>
using namespace std;
// Main Method declared here
int main()
{
int x;
cout<<"Enter any integer number: ";
cin>>x;
DecBin(x);
return 0;
}
// Here a function named DecBin is declared along with an integer variable, x
void DecBin(int x)
{
// Declare an array to store the resulting binary digits
int bindigit[32];
// counter for binary array
int kount = 0;
while (x > 0) {
// Store the remainder of each division in the declared array
bindigit[kount] = x % 2;
x = x / 2;
kount++;
}
// Loop to print the binary digits in reverse order
for (int j = i - 1; j >= 0; j--)
{
cout << bindigit[j];
}
}
// End of Program