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Ymorist [56]
3 years ago
11

Agile Software Development is based on Select one: a. Iterative Development b. Both Incremental and Iterative Development c. Inc

remental Development d. Linear Development
Computers and Technology
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

b. Both incremental and iterative development.

Explanation:

Agile software development is based on both incremental and iterative development. It is an iterative method to software delivery in which at the start of a project, software are built incrementally rather than completing and delivering them at once as they near completion.

Incremental development means that various components of the system are developed at different rates and are joined together upon confirmation of completion, while Iterative development means that the team intends to check again components that makes the system with a view to revising and improving on them for optimal performance.

Agile as an approach to software development lay emphasis on incremental delivery(developing system parts at different time), team work, continuous planning and learning, hence encourages rapid and flexible response to change.

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New trends, tools, and languages emerge in the field of web technology every day. Discuss the advantages of these trends, tools,
ikadub [295]

Answer:

Explanation:However, with the emergence of several new web development technologies, tools, frameworks, and languages in the last few years, it has now become quite .

7 0
3 years ago
Hexadecimal to denary gcse method
Anuta_ua [19.1K]

There are two ways to convert from hexadecimal to denary gcse method. They are:

  • Conversion from hex to denary via binary.
  • The use of base 16 place-value columns.

<h3>How is the conversion done?</h3>

In Conversion from hex to denary via binary:

One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.

Example - Find out the denary value of hex value 2D.

It will be:

2 = 0010

D = 1101

Put them them together and then you will have:

00101101

Which is known to be:

0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1

= 45 in denary form.

Learn more about hexadecimal from

brainly.com/question/11109762

#SPJ1

3 0
2 years ago
1
Dimas [21]

Answer:

C.  

Explanation:

Plato users

8 0
2 years ago
Given parameters b and h which stand for the base and the height of an isosceles triangle (i.e., a triangle that has two equal s
schepotkina [342]

Answer:

The area of the triangle is calculated as thus:

Area = 0.5 * b * h

To calculate the perimeter of the triangle, the measurement of the slant height has to be derived;

Let s represent the slant height;

Dividing the triangle into 2 gives a right angled triangle;

The slant height, s is calculated using Pythagoras theorem as thus

s = \sqrt{b^2 + h^2}

The perimeter of the triangle is then calculated as thus;

Perimeter = s + s + b

Perimeter = \sqrt{b^2 + h^2} + \sqrt{b^2 + h^2} +b

Perimeter = 2\sqrt{b^2 + h^2} + b

For the volume of the cone,

when the triangle is spin, the base of the triangle forms the diameter of the cone;

Volume = \frac{1}{3} \pi * r^2 * h

Where r = \frac{1}{2} * diameter

So, r = \frac{1}{2}b

So, Volume = \frac{1}{3} \pi * (\frac{b}{2})^2 * h

Base on the above illustrations, the program is as follows;

#include<iostream>

#include<cmath>

using namespace std;

void CalcArea(double b, double h)

{

//Calculate Area

double Area = 0.5 * b * h;

//Print Area

cout<<"Area = "<<Area<<endl;

}

void CalcPerimeter(double b, double h)

{

//Calculate Perimeter

double Perimeter = 2 * sqrt(pow(h,2)+pow((0.5 * b),2)) + b;

//Print Perimeter

cout<<"Perimeter = "<<Perimeter<<endl;

}

void CalcVolume(double b, double h)

{

//Calculate Volume

double Volume = (1.0/3.0) * (22.0/7.0) * pow((0.5 * b),2) * h;

//Print Volume

cout<<"Volume = "<<Volume<<endl;

}

int main()

{

double b, h;

//Prompt User for input

cout<<"Base: ";

cin>>b;

cout<<"Height: ";

cin>>h;

//Call CalcVolume function

CalcVolume(b,h);

//Call CalcArea function

CalcArea(b,h);

//Call CalcPerimeter function

CalcPerimeter(b,h);

 

return 0;

}

3 0
3 years ago
Assignment Background Video games have become an outlet for artists to express their creative ideas and imaginations and a great
Sphinxa [80]

Answer:

Code

import csv

def open_file():

try:

fname=input("Enter the filename to open (xxx.yyy): ")

fp=open(fname,encoding='utf-8')

return fp

except FileNotFoundError:

print("File not found! Please Enter correct filename with extension")

return open_file()

def read_file(fp):

fields = []

rows = []

csvreader = list(csv.reader(fp))

# skiped 1 row because it contains field name

fields = csvreader[0]

D1={}

D2={}

D3={}

# accessing the data of csv file

for line in csvreader[1:]:

name = line[0].lower().strip()

platform = line[1].lower().strip()

if line[2]!='N/A':

year = int(line[2])

else:

year=line[2]

genre = line[3].lower().strip()

publisher = line[4].lower().strip()

na_sales = float(line[5])

eur_sales = float(line[6])

jpn_sales = float(line[7])

other_sales = float(line[8])

global_sales=(na_sales+eur_sales+jpn_sales+other_sales)*1000000

if name not in D1:

D1[name]=[]

D1[name].append((name, platform, year, genre, publisher,global_sales))

if genre not in D2:

D2[genre]=[]

D2[genre].append((genre, year, na_sales, eur_sales,jpn_sales, other_sales, global_sales))

if publisher not in D3:

D3[publisher]=[]

D3[publisher].append((publisher, name, year, na_sales,eur_sales, jpn_sales, other_sales, global_sales))

# sorting

temp={}

# sorting keys

for Keys in sorted (D1) :

ls=D1[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D1=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D2) :

ls=D2[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D2=temp.copy()

temp={}

# sorting keys

for Keys in sorted (D3) :

ls=D3[Keys]

ls=sorted(ls,key=lambda x: x[-1]) # sorting values

temp[Keys]=ls

D3=temp.copy()

return D1,D2,D3

 

def main():

fp=open_file()

D1,D2,D3=read_file(fp)

# displaying the row_count in dictionaries

cnt=0

for d in D1:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D2:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

cnt=0

for d in D3:

cnt+=1

print(cnt)

print("\n\n=============================================\n\n")

 

 

if __name__=="__main__":

main()

Explanation:

see code and see output attached

7 0
3 years ago
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