<u>Throughput </u> is the actual speed of data transfer that is achieved between two nodes on a network and is always less than or equal to the data transfer rate.
<h3>What is used to transfer data packets between two or more networks?</h3>
A router is a networking device that forwards data packets between computer networks. Routers perform the traffic directing functions on the Internet.
<h3>What is throughput in data transfer?</h3>
In data transmission, network throughput is the amount of data moved successfully from one place to another in a given time period, and typically measured in bits per second (bps), as in megabits per second (Mbps) or gigabits per second (Gbps).
To learn more about Throughput , refer
brainly.com/question/25302150
#SPJ4
Answer:
def print_popcorn_time(bag_ounces):
if bag_ounces<3:
print("Too Small")
elif bag_ounces>10:
print("Too Large")
else:
total = 6*bag_ounces
print('{} seconds'.format(total))
Explanation:
Using Python programming language
The function is defined to accept a single parameter
Using a combination of if/elif/else statements, the approprite message is displayed when the function is called
Answer:
There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:
Let a,b be two numbers.
c=a+b;
print(c);
But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:
Let a be the variable where the input numbers are stored.
while(f==1)
{
printf(“Enter number”);
scanf(“Take number into the variable a”);
sum=sum+a;
printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);
scanf(“Take the input into the variable f”);
}
print(Sum)
Explanation:
hi there answer is given mar me as brainliest
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
Answer:
def recursive_func():
x = input("Are we there yet?")
if x.casefold() == 'Yes'.casefold():
return
else:
recursive_func()
recursive_func()
Explanation:
We define the required function as recursive_func().
The first line takes user input. The user input is stored in variable x.
The next line compares the user input to a string yes. The function executes the else block if the condition isn't met, that is a recursive call is executed.
IF condition returns the function. The string in variable X is compared to a string 'Yes'. the casefold() is a string function that ignores the upper/lower cases when comparing two strings. (This is important because a string 'yes' is not the same yes a string 'Yes' or 'YES'. Two equal strings means their cases and length should match).
