Which statement correctly describes an adaptive score?
2 answers:
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Answer:
Code is given below and output is attached in the diagram:
Explanation:
//Use this header file while using visual studio.
#include "stdafx.h"
//Include the required header files.
#include<iostream>
//Use the standard naming convention.
using namespace std;
//Define a namespace royaltyRates.
namespace royaltyRates
{
//Declare and initialize required named constants.
const double PAY_ON_DELIVERY_OF_NOVAL = 5000;
const double PAY_ON_PUBLISH_OF_NOVAL = 20000;
const double PER_ON_NET_PRICE_SECOND_OPTION =
0.125;
const double PER_ON_NET_PRICE_FIRST_4000 = 0.1;
const double PER_ON_NET_PRICE_OVER_4000 = 0.14;
};
//Start the execution of main() method.
int main()
{
//Declare and initialize the required variables
//which are going to be used to calculate
//royalities under each option.
float net_price;
int num_copies;
float royaltyUnderOption1, royaltyUnderOption2,
royaltyUnderOption3;
royaltyUnderOption1 = royaltyUnderOption2
= royaltyUnderOption3 = 0;
//Prompt the user to enter the net price of each
//novel.
cout << "Please enter the net price for each ";
cout << "copy of the novel : ";
cin >> net_price;
//Prompt the user to enter the estimated number
//of copies of the novels to be sold.
cout << "Please enter the estimated number ";
cout << "of copies to be sold : ";
cin >> num_copies;
//Display the required details and royalty
//calculated under option 1.
cout << "\n*** Option 1: ****" << endl;
cout << "Net price of each novel: $" << net_price;
cout << endl;
cout << "Estimated number of copies to be sold ";
cout << "is: " << num_copies << endl;
cout << "$";
cout << royaltyRates::PAY_ON_DELIVERY_OF_NOVAL;
cout << " is paid to author for the delivery of ";
cout << "the final manuscript and $";
cout << royaltyRates::PAY_ON_PUBLISH_OF_NOVAL;
cout << " is paid for the publication of ";
cout << "novel." << endl;
royaltyUnderOption1 =
royaltyRates::PAY_ON_DELIVERY_OF_NOVAL +
royaltyRates::PAY_ON_PUBLISH_OF_NOVAL;
cout << "Total amount of royalty under option 1 ";
cout << "is $" << royaltyUnderOption1 << endl;
//Display the required details and royalty
//calculated under option 2.
cout << "\n*** Option 2: ****" << endl;
cout << "Net price of each novel: $";
cout << net_price << endl;
cout << "Estimated number of copies to be sold ";
cout << "is: " << num_copies << endl;
royaltyUnderOption2 =
(royaltyRates::PER_ON_NET_PRICE_SECOND_OPTION *
net_price)* num_copies;
cout << "Total amount of royalty under option 2 ";
cout << "is $" << royaltyUnderOption2 << endl;
//Display the required details and royalty
//calculated under option 3.
cout << "\n*** Option 3: ****" << endl;
cout << "Net price of each novel: $" << net_price;
cout << endl;
cout << "Estimated number of copies to be sold ";
cout << "is: " << num_copies << endl;
//If the number of copies is greater than 4000.
if (num_copies > 4000)
{
//Total amount of royalty will be 10% of net
//price of first 4000 copies and 14 % of net
//price of copies sold over 4000.
royaltyUnderOption3 =
(royaltyRates::PER_ON_NET_PRICE_FIRST_4000 *
net_price) * 4000 +
(royaltyRates::PER_ON_NET_PRICE_OVER_4000 *
net_price) * (num_copies - 4000);
}
//Otherwise,
else
{
//Total amount of royalty will be 10% of net
//price of first 4000 copies.
royaltyUnderOption3 =
(royaltyRates::PER_ON_NET_PRICE_FIRST_4000 *
net_price) * num_copies;
}
cout << "Total amount of royalty under option 3 ";
cout << "is $" << royaltyUnderOption3 << endl;
//If the royalty under option 1 is greater than
//royalty under option 2 and 3, then option 1 is
//best option.
if (royaltyUnderOption1 > royaltyUnderOption2 &&
royaltyUnderOption1 > royaltyUnderOption3)
{
cout << "\nOption 1 is the best option that ";
cout << "author can choose." << endl;
}
//If the royalty under option 2 is greater than
//royalty under option 1 and 3, then option 2 is
//best option.
else if (royaltyUnderOption2 > royaltyUnderOption1
&& royaltyUnderOption2 > royaltyUnderOption3)
{
cout << "\nOption 2 is the best option that ";
cout << "author can choose." << endl;
}
//If the royalty under option 3 is greater than
//royalty under option 1 and 2, then option 3 is
//best option.
else
{
cout << "\nOption 3 is the best option that ";
cout << "author can choose." << endl;
}
//Use this command while using visual studio.
system("pause");
return 0;
}
Answer:
import string
dic = {}
book=open("book.txt","r")
# Iterate over each line in the book
for line in book.readlines():
tex = line
tex = tex.lower()
tex=tex.translate(str.maketrans('', '', string.punctuation))
new = tex.split()
for word in new:
if len(word) > 2:
if word not in dic.keys():
dic[word] = 1
else:
dic[word] = dic[word] + 1
for word in sorted(dic):
print(word, dic[word], '\n')
book.close()
Explanation:
The code above was written in python 3.
<em>import string </em>
Firstly, it is important to import all the modules that you will need. The string module was imported to allow us carry out special operations on strings.
<em>dic = {} </em>
<em>book=open("book.txt","r") </em>
<em> </em>
<em># Iterate over each line in the book</em>
<em>for line in book.readlines(): </em>
<em> </em>
<em> tex = line </em>
<em> tex = tex.lower() </em>
<em> tex=tex.translate(str.maketrans('', '', string.punctuation)) </em>
<em> new = tex.split() </em>
<em />
An empty dictionary is then created, a dictionary is needed to store both the word and the occurrences, with the word being the key and the occurrences being the value in a word : occurrence format.
Next, the file you want to read from is opened and then the code iterates over each line, punctuation and special characters are removed from the line and it is converted into a list of words that can be iterated over.
<em />
<em> </em><em>for word in new: </em>
<em> if len(word) > 2: </em>
<em> if word not in dic.keys(): </em>
<em> dic[word] = 1 </em>
<em> else: </em>
<em> dic[word] = dic[word] + 1 </em>
<em />
For every word in the new list, if the length of the word is greater than 2 and the word is not already in the dictionary, add the word to the dictionary and give it a value 1.
If the word is already in the dictionary increase the value by 1.
<em>for word in sorted(dic): </em>
<em> print(word, dic[word], '\n') </em>
<em>book.close()</em>
The dictionary is arranged alphabetically and with the keys(words) and printed out. Finally, the file is closed.
check attachment to see code in action.
Answer:
Explanation:
Farmers are always both directly and indirectly connected to each other
Their network is mostly strong
Networks become weak only on the edges (ends) of the river but doesn't completely dimnish
With the available network length, the center of river bank forms the strongest network of all and becomes a key player in defining the balance property of overall network
The network is very well structurally balanced and we can see that through the below image
20 miles 10 20 30 40 50
See attachment file for diagram
Considering the total length of river as 50miles and and the center of the whole length will be at 25th mile. From that point, if we consider a farmer will be be having friends for a length of 20miles both along upstream and downstream.
By this he'll be in friend with people who are around 80% of the total population. As me move from this point the integrity increases and this results in a highly balanced structural network.
