Response:
#section 1
def listofWordswitha(text):
''' This function outputs all words from a string that include the letter 'a' '''
tex = text.split()
new = []
#section 2
for word in tex:
for char in word:
if char == 'a':
new.append(word)
break
return new
Clarification:
#section 1
First, you need to establish your function and provide an argument representing the string that will be utilized.
It's beneficial to include a docstring that describes the function's purpose, which I've added.
Then, use the split method to break the string into a list. In conclusion, you create a list to store all the words that contain 'a'.
#section 2
The terminology chosen is specific to facilitate understanding.
FOR EVERY WORD inside the list and FOR EACH LETTER in the WORD.
IF a LETTER 'a' is found in the word, ADD that WORD to the NEW LIST.
The append function serves to incorporate new entries into the list.
A break statement is employed to avoid redundancy since some words might have multiple instances of 'a'. Thus, upon encountering a word containing 'a', it appends it and shifts to the next word.
Ultimately, the new list is returned.
I will utilize your inquiry to validate the code and provide the results for you.
Answer:
The required number of bits to address 8K words is 13.
Explanation:
Addressable words total 8000, where a word is defined as the smallest unit of memory that can be addressed.
These 8000 words can be accessed using units. To find the value of n corresponding to the number of words, you need to calculate
It's clear that 13 bits are necessary to address 8K words.