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nikklg [1K]
3 years ago
9

Taylor and Rory are hosting a party. They sent out invitations, and each one collected responses into dictionaries, with names o

f their friends and how many guests each friend is bringing. Each dictionary is a partial list, but Rory's list has more current information about the number of guests. Fill in the blanks to combine both dictionaries into one, with each friend listed only once, and the number of guests from Rory's dictionary taking precedence, if a name is included in both dictionaries. Then print the resulting dictionary.
Computers and Technology
1 answer:
disa [49]3 years ago
7 0

Answer:

Following are the code to this question:

def combine_guest(guest1, guest2):#defining a method combine_guest that accepts two dictionary

   guest2.update (guest1)#use dictionary guest2 that use update method to update guest2 dictionary

   return guest2#return guest2 dictionary values

Rory_guest= { "Ada":2, "Ben":3, "Dav":1, "Jo":3, "Charry":2, "Terry":1, "bob":4}#defining a dictionary and add value

Taylor_guest = { "Dav":4, "Nan":1, "bobert":2, "Ada":1, "Samantha":3, "Chr":5}#defining a dictionary and add value

print(combine_guest(Rory_guest,Taylor_guest))#calling the combine_guest method

Output:

{'Nan': 1, 'Samantha': 3, 'Ada': 2, 'bob': 4, 'Terry': 1, 'Jo': 3, 'Ben': 3, 'Dav': 1, 'Charry': 2, 'bobert': 2, 'Chr': 5}

Explanation:

In the code a method, "combine_guest" is defined, that accepts two dictionaries "guest1, guest2" inside the method, in which the "guest2" dictionary uses the update method to combine the value of the guest1 dictionary and use a return keyword to return guest2 values.

In the next step, two dictionaries are declared, that holds some values and use a print method to call the "combine_guest" method and prints its return values.  

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If you are trying to reduce the cost of college which of the following strategy is likely to save you the most money
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Write one for loop to print out each element of the list several things. then write another for loop
Kay [80]

Question:

Write one for loop to print out each element of the list several_things. Then, write another for loop to print out the TYPE of each element of the list several_things.

Answer:

The solution in python is as follows:

for element in several_things:

    print(element)

   

for element in several_things:

    print(type(element))

Explanation:

The solution assumes that the list several_things has already been initialized.

So, the rest of the code is explained as follows:

This line iterates through the list, several_things

for element in several_things:

This line prints each element

    print(element)

This line iterates through the list, several_things for the second time    

for element in several_things:

This line prints the type of each element

    print(type(element))

6 0
3 years ago
To mitigate the effects of most of the common network threats including disruption, destruction and disaster, companies are begi
almond37 [142]

Question:

To mitigate the effects of most of the common network threats including disruption, destruction and disaster, companies are beginning to migrate their servers, networking devices and data into professional datacenters. This is called

A) Colocation

B) SAAS

C) Peering

D) Clustering

E) Server Farming

Answer:

The correct answer is A) Colocation

Explanation:

Colocation as already defined is the voluntary relocation of all network facilities to a data centre so as to reduce the risks of disaster, disruption, destruction, intrusion whilst increasing security, flexibility and scalability at a lower cost.

As the world gets more digitized, datacentres are getting more patronage. The sales figures show this.

Cheers!

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Paha777 [63]

Answer:

30,33,37,18,23,34,15,38,40,17

To construct the AVL tree, follow these steps and diagrams are shown in the image:

• Add 30 to the tree as the root node. Then add 33 as the right child because of 33 is greater than 30 and AVL tree is a binary search tree.

• Then add 37 as the right child of 33. Here the balance factor of node 30 becomes 0-2 = -2, unbalanced.

Use RR rotation, make node 33 the root node, 30 as the left child of 33 and 37 as the right child of 33.

• Now add 18 as the left child of 30. And 23 as the right child of 18. Here the balance factor of 30 becomes 2-0 = 2. It’s unbalanced.

Use LR rotation, make 23 the parent of 18 and 30.  

• Now add 34 as the left child of 37 and 15 as the left child of 18. Add 38 as the right child of 37. And then add 40 as the right child of 38.

• Now adding 17 as the right child of 15 makes the tree unbalanced at 18.

Use LR rotation, make 17 as the parent of 15 and 18.

Explanation:

The balance factor of a node can be either 0,1 or -1. Else the tree is called unbalanced at the node.

If the inserted node is in the left subtree of the left subtree of the unbalance node, then perform LL rotation.

If the inserted node is in the right subtree of the right subtree of the unbalance node, then perform RR rotation.

If the inserted node is in the left subtree of the right subtree of the unbalance node, then perform RL rotation.

If the inserted node is in the right subtree of the left subtree of the unbalance node, then perform LR rotation.

7 0
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