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elena-14-01-66 [18.8K]
3 years ago
14

Which country was the first in the world to launch a mass media campaign in response to population growth?

Computers and Technology
1 answer:
Daniel [21]3 years ago
6 0
India is the correct answer 
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Productivity, rework, and technology impact are examples of which kind of software metric?
fgiga [73]

The correct answer should be process type of software metric. The process metrics are used to help in strategic decision making. The processes such as work products delivery, expended human hours and conformance to quality and schedule are all metrics in the process domain.

6 0
3 years ago
What is a router?
DIA [1.3K]

Answer:

A) a device that sends data to the receiving device

Explanation:

3 0
3 years ago
I need help please, thank you
swat32
Select all answers except computers, xbox, playstation 5, and mobile!

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8 0
3 years ago
Start with the following Python code. alphabet = "abcdefghijklmnopqrstuvwxyz" test_dups = ["zzz","dog","bookkeeper","subdermatog
trapecia [35]

Answer:

alphabet = "abcdefghijklmnopqrstuvwxyz"

test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

# From Section 11.2 of: # Downey, A. (2015). Think Python: How to think like a computer scientist. Needham, Massachusetts: Green Tree Press.

def histogram(s):

   d = dict()

   for c in s:

       if c not in d:

           d[c] = 1

       else:

           d[c] += 1

   return d

#Part 1 Write a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.

def has_duplicates(stringP):

   dic = histogram(stringP)

   for key,value in dic.items():

       if value>1:

           return True

   return False

# Implement has_duplicates by creating a histogram using the histogram function above. Write a loop over the strings in the provided test_dups list.

# Print each string in the list and whether or not it has any duplicates based on the return value of has_duplicates for that string.

# For example, the output for "aaa" and "abc" would be the following. aaa has duplicates abc has no duplicates Print a line like one of the above for each of the strings in test_dups.

print("***Implementation of has_duplicates fuction***")

for sTr in test_dups:

   if has_duplicates(sTr):

       print(sTr+": has duplicates")

   else:

       print(sTr+": has no duplicates")

#Part 2 Write a function called missing_letters that takes a string parameter and returns a new string with all the letters of the alphabet that are not in the argument string.

#The letters in the returned string should be in alphabetical order. Your implementation should use a histogram from the histogram function. It should also use the global variable alphabet.

#It should use this global variable directly, not through an argument or a local copy. It should loop over the letters in alphabet to determine which are missing from the input parameter.

#The function missing_letters should combine the list of missing letters into a string and return that string.

def missing_letters(sTr):

   missingLettersList = []

   dic = histogram(sTr)

   for l in alphabet:

       if l not in dic:

           missingLettersList.append(l)

   missingLettersList.sort()

   return "".join(missingLettersList)

#Write a loop over the strings in list test_miss and call missing_letters with each string. Print a line for each string listing the missing letters.

#For example, for the string "aaa", the output should be the following. aaa is missing letters bcdefghijklmnopqrstuvwxyz

#If the string has all the letters in alphabet, the output should say it uses all the letters.

#For example, the output for the string alphabet itself would be the following. abcdefghijklmnopqrstuvwxyz uses all the letters

#Print a line like one of the above for each of the strings in test_miss.

print("\n***Implementation of missing_letters fuction***")

for lTm in test_miss:

   sTr = missing_letters(lTm.replace(" ",""))

   if sTr!="":

       print(lTm+" is missing letters "+sTr)

   else:

       print(lTm +" uses all the letters")

3 0
3 years ago
We may think of relationships in the E/R model as having keys, just as entity sets do. Let R be a relationship among the entity
BaLLatris [955]

Explanation:

1

R is many - many;

in this case we have that E1 is not what should be used to determine E2 and in this same way, W2 is not what should determine E1. So non of these can be a key on its own.

2.

R is many-one from E1 to E2

in this case there are 2 tuples (f1,f2) and (e1,e2) which have to be the same if they have the same key attribute for E1. All of them are the same so all the pairs are the same.

3.

R is many-one from E2 to E1:

the explanation for this is is almost the same with that of b. we apply the same logic. since for all keys, all pairs and sets are the same.

4.

R is one-one:

E1 is what determines E2 and likewise, E2 is what determines E1. irrespective of the key that is used, key K is going to the same value of n.

8 0
3 years ago
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