The answer would be Assembly language as assembly language is not used for web development, but rather is a low level programming language.
Please mark branliest if this helped!!
Answer:
The correct answer is: Option d: Put Quotation marks around the string.
Explanation:
The names of things or places or cars are usually in the form of words or sentences. The words or sentences are stored in string datatype.
Quotation marks are used to deal with strings in programming languages.
Hence,
The correct answer is: Option d: Put Quotation marks around the string.
Depends really. Have you got prior coding experience? The semantics of certain languages are much easier to pick up if you've had prior experience, but it might seem impossible to someone who has minimal experience.
Assuming that you're no older than 15 or 16, I'm going to suggest Python. It's a simple, high-level language that's easier to understand than most languages. If you think you've got enough experience to quickly understand and pick up things, then I'd probably say R is a good language to start with. It's very well-suited for AI and tends to be a favourite for AI researchers.
Answer:
C: Security
Explanation:
Communications and information systems principles need to be, among other things, secure. They need to be able to protect sensitive information from those who intentionally not need to know. Some incident information like voice, networks, and data, are very sensitive and thus, should be secure to the right levels and should comply with privacy laws and data protection.
Answer:
See explaination for the code
Explanation:
def wordsOfFrequency(words, freq):
d = {}
res = []
for i in range(len(words)):
if(words[i].lower() in d):
d[words[i].lower()] = d[words[i].lower()] + 1
else:
d[words[i].lower()] = 1
for word in words:
if d[word.lower()]==freq:
res.append(word)
return res
Note:
First a dictionary is created to keep the count of the lowercase form of each word.
Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.
Finally the res list is appended.
