Answer:
Whether a traffic light is green, yellow or red
Explanation:
Boolean variables are variables that can either take one out of two options at any given instance.
Analyzing the given options
1. Elevator:
Possible Directions = Up or Down; That's two possible values
But it can only move in one direction at a given instance.
<em>This can be represented using Boolean</em>
2. Traffic Light:
Possible Lights = Green, Yellow or Red
That's three options.
<em>This can be represented using Boolean</em>
<em />
The last two options can be represented using Boolean because they have just two possible values
<em>Hence, option (B) answers the question</em>
Well hypothetically I've seen a lot of cars but like Bentleys Lamborghinis Ferraris and Rolls Royce's but it'll probably hypothetically have to be my mom 2004 Toyota camery
Answer:
The correct answer is:
"joining a Python developer forum and posting a question to the forum to solicit feedback"
Explanation:
Learning a new skill involves a lot of research and study especially learning a new programming language.
The syntax and commands have to be understood first.
Now if Sam has to implement a particular feature, the easiest and less time-consuming way is that he post his query on a Python language forum as there might be better and expert programmer that might help
Hence,
The correct answer is:
"joining a Python developer forum and posting a question to the forum to solicit feedback"
Answer:
There's a parking lot that is 600m² big. The lot must be able to hold at least 3 buses and 10 cars.
Each car takes up 6m² and each bus takes up 30m².
However, there can only be 60 vehicles in the lot at any given time.
The cost to park in the lot is $2.50 per day for cars and $7.50 per day for buses. The lot must make at least $75 each day to break even.
What is a possible car to bus ratio that would allow the lot to make profit?
Answer:
def most_frequent_letter():
file = open("words","r")
dWords = {}
for line in file:
line = line.rstrip()
words = line.split(" ")
for word in words:
counts = {}
for c in word:
counts[c] = counts.get(c, 0) + 1
dWords[word] = max(counts, key=counts.get)
return dWords
print(most_frequent_letter())
Explanation:
the file used was words in txt format and its contents are as follows:
hello aamir jan khan
parallelogram abdullah
anaconda ali
pycharm notebook
