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3 years ago

To delete a comment, right click the comment and then select the ____ command.

Computers and Technology

2 answers:

aleksandrvk [35]3 years ago

8 0

delete command is your answer i tink

jekas [21]3 years ago

3 0

On the Review tab, in the Comments group, click the Delete command.

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In this question, we give two implementations for the function: def intersection_list(lst1, lst2) This function is given two lis

castortr0y [4]

Answer:

see explaination

Explanation:

a)Worst Case-time complexity=O(n)

def intersection_list(lst1, lst2):

lst3 = [value for value in lst1 if value in lst2]

return lst3

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele) # adding the element

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele) # adding the element

print(intersection_list(lst1, lst2))

b)Average case-time complexity=O(min(len(lst1), len(lst2))

def intersection_list(lst1, lst2):

return list(set(lst1) & set(lst2))

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele)

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele)

print(intersection_list(lst1, lst2))

6 0

3 years ago

Suppose users share a 1-Gbps link. Also, suppose each user requires 200 Mbps when transmitting, but each user only transmits 30

V125BC [204]

Answer:

The answer is "5 users and 1 block".

Explanation:

In Option a:

Bandwidth total $= 1-Gbps \times 1000$

$= 1,000 \ Mbps$

Any User Requirement $= 200 \ Mbps$

The method for calculating the number of approved users also is:

Now, calculate the price of each person for overall bandwidth and demands,

$\text{Sponsored user amount} = \frac{\text{Bandwidth total}}{\text{Each user's requirement}}$

$=\frac{1000}{200}\\\\=\frac{10}{2}\\\\= 5 \ users$

In Option b:

$\text{blocking probability} = \frac{link}{\text{1-Gbps} = 10^9 \frac{bits}{sec}}$

$\ let = 0.05 = \frac{100}{20} \\\\\text{blocking probability} = \frac{ 200 \times 10^6}{\frac{100}{20}}$

$= \frac{ 200 \times 10^6 \times 20 }{100}\\\\= \frac{ 2 \times 10^6 \times 20 }{1}\\\\= 40 \times 10^6 \\\\$

mean user $= 25 \times \frac{1}{20} \\\\$

$= 1.25$

max user $= \frac{10^9}{40 \times 10^6} \\\\$

$= \frac{10^9}{4 \times 10^7} \\\\ = \frac{10^2}{4} \\\\ = \frac{100}{4} \\\\= 25 \\\\ =\ \ 1 \ \ block \\\\$

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The right answer is A

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2 years ago

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