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hammer [34]
1 year ago
9

The variable grade can have any real number value from 0 to 100. Ask the user to enter a grade in numerical form. Write an if-el

seif-else statement that displays the letter grade (any format) corresponding to a numerical grade in an appropriately formatted output statement. Use the standard 10-point grading scale:  a: A: 90 ≤ grade  a: B: 80 ≤ grade < 90  a: C: 70 ≤ grade < 80  a: D: 60 ≤ grade < 70  a: F: grade < 60
Computers and Technology
1 answer:
Alborosie1 year ago
8 0

Answer:

See explaination

Explanation:

#include <iostream>

using namespace std;

int main()

{

float marks;

cout<<"Enter your grade value :";

cin>>marks;

if(marks>=90)

cout<<"Your grade is 'A'";

else if(marks>=80&& marks<90)

cout<<"Your grade is 'B'";

else if(marks>=70&& marks <=80)

cout<<"Your grade is 'C'";

else if(marks>=60&& marks <70)

cout<<"Your grade is 'D'";

else

cout<<"Your grade is 'F'";

}

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What would make this comparison statement False? Complete with the appropriate relational operator. "G" _____= "G"
My name is Ann [436]

Equality and Relational Operators

For the statement to return false, you can simply use the "not equal to" equality operation. The full symbol of this operation is '!=', disregarding the quotes.

<u>Examples:</u>

  • [1 != 1]  would produce FALSE. Translation: 1 <u>does not equal</u> 1?
  • [1 == 1]  would produce TRUE. Translation: 1 <u>does</u> 1?
  • ["G" != "G]  would produce <u>FALSE</u>. Translation: "G" <u>does not equal</u> "G"?

CONCLUSION: Use "!=".

8 0
2 years ago
When you complete a form online with your name, address, and credit card information in order to make a purchase, you are giving
mafiozo [28]

Answer:

Structured

Explanation:

A data dictionary can be defined as a centralized collection of information on a specific data such as attributes, names, fields and definitions that are being used in a computer database system.

In a data dictionary, data elements are combined into records, which are meaningful combinations of data elements that are included in data flows or retained in data stores.

This ultimately implies that, a data dictionary found in a computer database system typically contains the records about all the data elements (objects) such as data relationships with other elements, ownership, type, size, primary keys etc. These records are stored and communicated to other data user when required or needed.

A relational database can be defined as a type of database that is structured in a manner that there exists a relationship between its elements.

Hence, when a user completes a form online by providing his or her name, address, and credit card information in order to make a purchase, he or she is giving the company structured data (pre-defined and formatted to a set structure) that can be used to make up (populate) a relational database or spreadsheet.

A spreadsheet can be defined as a file or document which comprises of cells in a tabulated format (rows and columns) typically used for formatting, arranging, analyzing, storing, calculating and sorting data on computer systems.

3 0
1 year ago
If s=abcd is a string defined over Σ = {a,bc,d}then reverse of s is dcba.<br> Δ True<br> Δ False
Minchanka [31]

Answer:

True

Explanation:

If s=abcd is a string defined over {a,b,c,d}, it corresponds to a regular expression which can be represented using a finite automata. Then the reverse of the string essentially corresponds to another finite automata where the starting state becomes the accepting state and vice versa.Moreover all the directions of state transitions will be reversed for each of the transitions in the original automata.

With these modifications, the new finite automata will accept a string which is reverse of the original string ,namely, dcba and this string will ne part of the reverse language.

7 0
1 year ago
Write a while loop that prints user_num divided by 2 until user_num is less than 1. The value of user_num changes inside of the
WITCHER [35]

Answer:

The code to this question can be defined as follows:

Code:

#include <stdio.h> //defining header file

int main() //defining main method

{

float user_num; //defining float variable

printf("Enter a value: "); //message

scanf("%f",&user_num); //input value from user end

printf("%f, ",user_num); //print value

while (user_num>1) //loop to calculte value

{

user_num=user_num/2; //diving value

printf("%f, ",user_num); //print value.

}

   return 0;

}

Output:

Enter a value: 20

20.000000, 10.000000, 5.000000, 2.500000, 1.250000, 0.625000,  

Explanation:

Description of the code as follows:

  • First, a float variable "user_num" is declared, in which we take input from the user-end.
  • In the next step, a while loop is declared, that uses the above variable to calculates its value.
  • Inside the loop variable "user_num" divide its value by 2 and holds its calculated value, to print its value the "printf" method is used that prints its value.  
8 0
2 years ago
5.14 ◆ Write a version of the inner product procedure described in Problem 5.13 that uses 6 × 1 loop unrolling. For x86-64, our
KengaRu [80]

Answer:

(a) the number of times the value is performs is up to four cycles. and as such the integer i is executed up to 5 times.  (b)The point version of the floating point can have CPE of 3.00, even when the multiplication operation required is either 4 or 5 clock.

Explanation:

Solution

The two floating point versions can have CPEs of 3.00, even though the multiplication operation demands either 4 or 5 clock cycles by the latency suggests the total number of clock cycles needed to work the actual operation, while issues time to specify the minimum number of cycles between operations.

Now,

sum = sum + udata[i] * vdata[i]

in this case, the value of i performs from 0 to 3.

Thus,

The value of sum is denoted as,

sum = ((((sum + udata[0] * vdata[0])+(udata[1] * vdata[1]))+( udata[2] * vdata[2]))+(udata[3] * vdata[3]))

Thus,

(A)The number of times the value is executed is up to 4 cycle. And the integer i performed up to 5 times.

Thus,

(B) The floating point version can have CPE of 3.00, even though the multiplication operation required either 4 or 5 clock.

3 0
1 year ago
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