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malfutka [58]
2 years ago
5

Write a program that lets a user enter N and that outputs N! (N factorial, meaning N*(N-1)*(N-2)*..\.\*2*1). Hint: Initialize a

variable total to N (where N is input), and use a loop variable i that counts from total-1 down to 1. Compare your output with some of these answers: 1:1, 2:2, 3:6, 4:24, 5:120, 8:40320.
Computers and Technology
1 answer:
Mrac [35]2 years ago
6 0

Answer:

N = int(input("Enter a number: "))

total = N

for i in range(N-1, 1, -1):

   total *= i

   

print(total)

Explanation:

Ask the user for the input, N

Set the total N at the beginning

Create a for loop that iterates from N-1 to 1 inclusively. Inside the loop, multiply each number in that range and add the result to the total.

When the loop is done, print the total to see the result

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Object-oriented development could potentially reduce the time and cost of writing software because: Group of answer choices a) i
spayn [35]

Answer:

b) objects are resuable

Explanation:

In OOP there's code reuse where a method or any other body of code is defined once and called or reused severally.

3 0
2 years ago
Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combinin
Dmitry_Shevchenko [17]

Answer:

Follows are the code to this question:

#include<iostream>//defining header file

using namespace std;// use package

int main()//main method

{

int red,green,blue,x;//declaring integer variable

cin>> red >>green>>blue;//use input method to input value

if(red<green && red<blue)//defining if block that check red value is greater then green and blue  

{

x = red;//use x variable to store red value

}

else if(green<blue)//defining else if block that check green value greater then blue  

{

x= green; //use x variable to store green value

}

else//defining else block

{

x=blue;//use x variable to store blue value

}

red -= x;//subtract input integer value from x  

green -=x; //subtract input integer value from x

blue -= x;//subtract input integer value from x

cout<<red<<" "<<green<<" "<<blue;//print value

return 0;

}

Output:

130 50 130

80 0 80

Explanation:

In the given code, inside the main method, four integers "red, green, blue, and x" are defined, in which "red, green, and blue" is used for input the value from the user end. In the next step, a conditional statement is used, in the if block, it checks red variable value is greater than then "green and blue" variable. If the condition is true, it will store red variable value in "x", otherwise, it will goto else if block.

  • In this block, it checks the green variable value greater than the blue variable value. if the condition is true it will store the green variable value in x variable.
  • In the next step, else block is defined, that store blue variable value in x variable, at the last step input variable, is used that subtracts the value from x and print its value.      
5 0
2 years ago
Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0
2 years ago
Type the correct answer in the box. Spell all words correctly.
kvasek [131]

Answer:

Presentations?

Explanation:

6 0
2 years ago
Sea level is the average level of the sea between high and low tide. It is used as a reference point for measuring elevation, or
SSSSS [86.1K]

Answer:

-8 and 735

Explanation:

Since the sea is a body of water (fluid) it takes the same properties as a fluid does which means filling an area and finding a leveled point. This is why we use the sea as a starting elevation point for Earth's land and it is given a numerical value of 0. Since it starts at 0 any land above Sea Level is in the positives while land below Sea Level is in the negatives. Therefore Since New Orleans is 8 feet below sea level its numerical value elevation is -8. The highest point in Chicago is 735 feet above sea level which makes its numerical value 735

6 0
2 years ago
Read 2 more answers
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