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NISA [10]
3 years ago
10

How does a hub-and-spoke network increase operational efficiency?

Computers and Technology
1 answer:
tatuchka [14]3 years ago
7 0
<span>Significantly fewer routes are needed to serve the network. This is because the number of pairings in a P2P network increases at a greater rate than the increase in nodes. (For those familiar with Big-O notation, it’s O(n^2) ).</span>
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Select each of the steps involved in creating a table in a presentation.
klio [65]

Answer:

huh? Making your presentation more interesting through the use of multimedia can help to improve the audience's focus. PowerPoint allows you to use images, audio and video to have a greater visual impact. These visual and audio cues may also help a presenter be more improvisational and interactive with the audience.

7 0
3 years ago
PLEASE HELP QUICK ASAP NO LINKS PYTHON CODING
Mademuasel [1]

Answer:

its for the movement

Explanation:

6 0
3 years ago
Read 2 more answers
) Suppose algorithm A takes 10 seconds to handle a data set of 1000 records. Suppose the algorithm A is of complexity O(n2). Ans
Mice21 [21]

Answer:

i) The time taken for 1500 records = 15 seconds.

ii) The time taken for 1500 records = 50 seconds.

Explanation:

A is an O(n) algorithm.

An algorithm with O(n) efficiency is described as a linearly increasing algorithm, this mean that the rate at which the input increases is linear to the time needed to compute that algorithm with n inputs.

From the question, it is given that for 1000 records, the time required is: 10 seconds.

Algorithm time taken is O(n)

Hence,

1) For 1,500 records

=> 10/1000 = x/1500

=> 10x1500/1000 = x

x = 15 seconds

Thus, the time taken for 1500 records = 15 seconds.

2) For 5,000 records

=> 10/1000 = x/5000

=> 10x5000/1000 = x

x = 50 seconds

Thus, the time taken for 1500 records = 50 seconds.

8 0
3 years ago
We love him, because he ."
Andre45 [30]
We love him, because he “is everything I need.” I’m kinda confused on the answer lemme know if you need help
7 0
2 years ago
Python
andreyandreev [35.5K]

Answer:

<h2>1)   </h2>

h = 19  # h is already assigned a positive integer value 19

i = 1  # used to calculate perfect square

q = 0   # stores the sum of the perfect squares

while q < h:  

# this loop executes until value of number of perfect squares is less than h

   q = q + (i*i)  # to calculate the sum of perfect squares

   i = i + 1  # increments i by 1 at every iteration

print(q) # displays the sum of perfect squares

Output:

30

Explanation:

Here we see that the sum of the perfect square is printed in the output i.e. 30. If we want to assign value 4 to q then we should alter the code as  following:  

h = 19

i = 1

sum_of_ps = 0

q=0

while sum_of_ps < h:

   sum_of_ps = sum_of_ps + (i*i)

   q = q + 1

   i = i + 1  

print(q)  

If you want to take input from the user and also display the perfect squares in output you can use the following code:

h = int(input('Enter the value of h: ')) #takes value of h from user

i = 1

sum_of_ps = 0

q=0

while sum_of_ps < h:

   sum_of_ps = sum_of_ps + (i*i)

   print(i*i) #displays the perfect squares of h

   q = q + 1

   i = i + 1    

print(sum_of_ps)   #displays sum of perfect squares

print(q) #displays number of perfect squares whose value is less than h

Output:

1

4

9

16

30

4

<h2>2)</h2>

# k and m are already assigned positive integers 10 and 40

k = 10  

m = 40  

i = 1  # i is initialized to 1

q = 0  # q stores the number of perfect squares between k and m

while i*i < m:  # loop computes perfect square of numbers between k and m

   if i*i > k:          

       q = q + 1  # counts the number of perfect squares

   i = i + 1  #increments the value of i at each iteration

print (q) # prints the no of perfect squares between k and m

If you want to display the perfect squares too use , add print(i*i) statement after if statement  if i*i > k:        

Output:

3

<h2>3)</h2>

def Fibonacci(n):  # method Fibonacci which has an integer parameter n

   if n <= 1:  # if the value of n is less than or equals to 1

       result = n  # stores the value of n into result

   elif n > 1:  # else if the value of n is greater than 1

       result = Fibonacci(n-1) + Fibonacci(n-2)  

# calls Fibonacci method recursively to associate nth value of Fibonacci #sequence with variable result

   return result     #returns the final value stored in result variable

print(Fibonacci(8))

#calls Fibonacci function and passes value 8 to it

Output:

21

8 0
3 years ago
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