Answer: a. 0.4667
b. 0.4667 and C 0.0667
Explanation:
Given Data:
N = population size (10)
n = random selection (2)
r = number of observations = 7
Therefore
f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )
When y = 1
f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 2
f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 0
f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )
= 1 / 15
= 0.0667
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
Answer:
A wheelbarrow, a bottle opener, and an oar are examples of second class levers
Answer:
class TriangleNumbers
{
public static void main (String[] args)
{
for (int number = 1; number <= 10; ++number) {
int sum = 1;
System.out.print("1");
for (int summed = 2; summed <= number; ++summed) {
sum += summed;
System.out.print(" + " + Integer.toString(summed));
}
System.out.print(" = " + Integer.toString(sum) + '\n');
}
}
}
Explanation:
We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.
