Calculate The Average of Grades Instructions:
Please read the following problem carefully. You will then logon on to https://www.draw.io/to create a professional diagram. You will put all titles, labels, and save your work using the information below. Please follow the directions:
Destini would like to get a better understanding of her grades in all of her college courses before Spring Break. Instead of using a calculator and paper to calculate her grade, she decided to design a program. She will design a program that will ask her to enter her course name and the number of grades in her grade book. The program will then Read the number of Grades based on what was entered by her, Add up all the Grades, Calculate the Average, and Display Course Name and the Average to Screen. It is important to consider that the number of grades will be different for each course. Using Repetition Control, please design this program.
1. Please Create Flowchart using Draw.10 and Simple Flowchart symbols only (Draw.IO)
2. Pseudocode.
Your unique Flowchart must have the following (See Example Here):
A. Name, Date, and class Name [Top Left of Flowchart)
B. A Title of the Flowchart in Bold [Centered at the top of your flowchart)
C. A Brief Summary of your Flowchart (2 to 3 short sentences describing your flowchart) [Left of Flowchart]
Answers
Answer:
The pseudocode is as follows:
Input coursename, numgrades
count = 1; totalgrades = 0
while count <= numgrades:
input grade
totalgrade+=grade
count++
average = totalgrade/count
print(coursename)
print(average)
Explanation:
The solution is as follows:
(1) See attachment for flowchart
(2) See answer section for pseudocode
Explanation
Input coursename and number of grades
Input coursename, numgrades
Initialize count of grades input by the user to 1 and the sum of all grades to 0
count = 1; totalgrades = 0
This loop is repeated while count of grades input by the user is less than or equal to the numgrades
while count <= numgrades:
Input grade
input grade
Add grades
totalgrade+=grade
Increase count by 1
count++
End of loop
Calculate average
average = totalgrade/count
Print coursename
print(coursename)
Print average
print(average)
C. Summary of the flowchart
The flowchart gets coursename and the number of grades from the user. Then it gets the score of each grade, add them up t calculate the average of grades.
Lastly, the course name and the average grades is printed
Related Questions
Log onto the Internet and use a search engine to find three Web sites that can be models for the new site. (At least one should sell sporting goods.) For each site you selected, list the URLs in your document. Tell why you chose them. Navigate to the three sites you choose and take notes about at least three things you like and don’t like about the sites.
Answers
Please see the following file for your required answer, thank you.
An integrated circuit RAM chip has a capacity of 512 words of 8 bits each where the words are organized using a one-dimensional layout.a) How many address lines are there on the chip?b) How many such chips would be needed to construct a 4Kx16 memory?c) How many address and data lines needed for a 4Kx16 memory?#addresss lines = #data lines =
Answers
Answer:
A. 9
B. 16
C. Number of addresses = 12 number of data lines = 16
Explanation:
The capacity of this chip is 512 x 8
That is 512 words and these words have 8 bits
A. The number of the address lines would be
log2(512) = 2⁹
So the answer is 9
B. The total number of chips required
= 512 x 8 = 4096
(4096 x 2)/ 512 = 16
C. Number of address = log4092 = 2¹²
= 12
The number of data lines = 8x2 = 16
Thank you!
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Answers
Answer:
http://www.hackshop.org/levels/basic-arduino/hack-the-chromebook
Explanation:
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Answers
Answer:
10.b
11.c
12.c
13.a
14.d
15.b
16.c
Explanation:
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Suppose a byte-addressable computer using set-associative cache has 2 16 bytes of main memory and a cache size of 32 blocks and each cache block contains 8 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the size of the tag, set, and offset fields
Answers
Answer:
Following are the responses to these question:
Explanation:
The cache size is 2n words whenever the address bit number is n then So, because cache size is 216 words, its number of address bits required for that cache is 16 because the recollection is relational 2, there is 2 type for each set. Its cache has 32 blocks, so overall sets are as follows:
[tex]\text{Total Number of sets raluired}= \frac{\text{Number of blocks}}{Associativity}[/tex]
[tex]=\frac{32}{2}\\\\ =16\\\\= 2^4 \ sets[/tex]
The set bits required also are 4. Therefore.
Every other block has 8 words, 23 words, so the field of the word requires 3 bits.
For both the tag field, the remaining portion bits are essential. The bytes in the tag field are calculated as follows:
Bits number in the field tag =Address Bits Total number-Set bits number number-Number of bits of words
=16-4-3
= 9 bit
The number of bits inside the individual fields is therefore as follows:
Tag field: 8 bits Tag field
Fieldset: 4 bits
Field Word:3 bits
Create a map using the Java map collection. The map should have 4 entries representing students. Each entry should have a unique student ID for the key and a student name for the element value. The map content can be coded in directly, you do not have to allow a user to enter the map data. Your program will display both the key and the value of each element.
Answers
Answer:
In Java:
import java.util.*;
public class Main{
public static void main(String[] args) {
Map<String, String> students = new HashMap<>();
students.put("STUD1", "Student 1 Name");
students.put("STUD2", "Student 2 Name");
students.put("STUD3", "Student 3 Name");
students.put("STUD4", "Student 4 Name");
for(Map.Entry m:students.entrySet()){
System.out.println(m.getKey()+" - "+m.getValue()); }
}}
Explanation:
This creates the map named students
Map<String, String> students = new HashMap<>();
The next four lines populates the map with the ID and name of the 4 students
students.put("STUD1", "Student 1 Name");
students.put("STUD2", "Student 2 Name");
students.put("STUD3", "Student 3 Name");
students.put("STUD4", "Student 4 Name");
This iterates through the map
for(Map.Entry m:students.entrySet()){
This prints the details of each student
System.out.println(m.getKey()+" - "+m.getValue()); }