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Deffense [45]
2 years ago
15

Three groups of students are given study outlines for 6 weeks. One group studies 2 hours a night, a second group studies 1 hour

a night, and the third group only studies 30 minutes. Their final exam grade will be at the end of the 6 weeks. What is the constant ?
Engineering
1 answer:
katrin2010 [14]2 years ago
8 0

Answer:

The constant here is the study outline

Explanation:

In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.

Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.

NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.

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Staying hydrated at all times
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Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with
ehidna [41]

Answer:

Um...

Explanation:

This is what I like to see teachers giving out.

7 0
3 years ago
For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our w
sergiy2304 [10]

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

8 0
3 years ago
O que é necessário se conhecer para a determinação de uma força concentrada em um ponto?
skelet666 [1.2K]

Answer: its c

Explanation:

7 0
3 years ago
A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in t
uysha [10]

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

8 0
2 years ago
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