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Fynjy0 [20]
3 years ago
5

Based on three scores that the user inputs, display the average of the score and the letter grade that is assigned fort he test

score average. The grading scheme is as follows:
// 90 - 100 A
// 80 - 89 B
// 70 - 79 C
// 60 - 69 D
// <60 F
return 'F';
Your display should look like:
Your average test score is 89.9. You earned a B.
1 public void calculateDisplayLetterGrade (double scorel, double score2, double score3)
2 {
3
4 }
5
Computers and Technology
1 answer:
Korolek [52]3 years ago
5 0

I used python function to write the code.

def averageScore(x,y,z):

    s = x + y + z

    avg = s/3

   if avg >= 90:

              return f'Your average test score is {avg} You earned a A.'

    elif avg >=80 and avg < 90 :

                return f'Your average test score is {avg} You earned a B.'

   elif avg >=70 and avg < 80 :

                 return f'Your average test score is {avg} You earned a .C'

      elif avg >=60 and avg < 70 :

                  return f'Your average test score is {avg} You earned a D.'

      else:

                    return f'Your average test score is {avg} You earned a F.'

print(averageScore(98, 90, 80))

Python function is used to write the code where x, y and z are the argument (user inputs) of the function. Then the inputs are summed and the average is gotten and stored with the variable, avg.  

If the average is greater than or equal to 90 the program will give the appropriate score and response(grade). It does the same if the user scored 80 - 89, 70 - 79, 60 - 69 and less than 60.

Finally. the function is called with the user input(argument).

Note the bolded values in the code are keywords in python.

read more: brainly.com/question/14191443?referrer=searchResults

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Suppose we compute a depth-first search tree rooted at u and obtain a tree t that includes all nodes of g.
Temka [501]

G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

<h3>What are DFS and BFS?</h3>

An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.

The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.

First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.

If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.

In such a case:

- in the DFS tree, one of u or v, should be a prototype of the other.

- in the BFS tree, u and v can differ by only one level.

Since, both DFS-tree and BFS-tree are the very tree T,

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This means that the border joining them must be in T.

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In the two-part of evidence:

Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

The complete question is:

We have a connected graph G = (V, E), and a specific vertex u ∈ V.

Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.

Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.

Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)

To learn more about  DFS and BFS, refer to:

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3. Excessive Use of the Internet. Solutions are:

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<h3>What is Digital Literacy?</h3>

Digital literacy can be defined as the ability or skills needed to be able to access information and communicate via internet platforms, mobile devices, or social media.

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Learn more about digital literacy on:

brainly.com/question/14242512

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