Answer:
D
Explanation:
ensuring project end on time through carefully planning and organizing
Answer:
See explaination and attachment.
Explanation:
Iteration method is a repetitive method applied until the desired result is achieved.
Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below
x = pi(x)
Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).
Similarly for second, thrid and so on. approximation
x2 = pi(x1)
x3 = pi(x2)
x4 = pi(x3)
xn = pi(xn-1).
please go to attachment for the step by step solution.
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
Answer:
It’s so we can then understand other cultures and other ways that people do things or we can also be prepared for what a country that we have not been to yet is going to relatively be like. This can also help with vacation planning.
Explanation:
it can also help us understand different people from different cultures.
Answer:
insert (array[] , value , currentsize , maxsize )
{
if maxsize <=currentsize
{
return -1
}
index = currentsize-1
while (i>=0 && array[index] > value)
{
array[index+1]=array[index]
i=i-1
}
array[i+1]=value
return 0
}
Explanation:
1: Check if array is already full, if it's full then no component may be inserted.
2: if array isn't full:
- Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
- Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
- assign new worth to the position that is next to the known position of initial smaller component.
