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AfilCa [17]
3 years ago
8

PlZZZZZZZ help will give Brainliest

Computers and Technology
2 answers:
melamori03 [73]3 years ago
7 0
I believe it is called “Transition”
dalvyx [7]3 years ago
3 0

Answer:

Phone

Explanation:

You might be interested in
Write a basic notation for
Licemer1 [7]

Answer:

a = (b + c)/(2 * c)

z = x/(y + c)

c = (9 * c + 32)/5

Explanation:

Required

The expression in basic

To do this, we use () to group items, / as divide and * to combine factors

So, we have:

(a)\ a = \frac{(b + d)}{2c}

In basic, it is:

a = (b + c)/(2 * c)

(b)\ z = \frac{x}{y + c}

In basic, it is:

z = x/(y + c)

(c)\ c = \frac{9c + 32}{5}

c = (9 * c + 32)/5

4 0
3 years ago
A/an<br> controls the flow of data between a computer and the network.
strojnjashka [21]

Answer:

The NIC (Network Interface Card)  is the computer network component that controls the flow of data between a computer and the network.

Explanation:

5 0
3 years ago
Create a series called 'directors' that contains each director's name and his or her average rating. Print out the type of your
sasho [114]

Using the computational language in python it is possible to write a code that we will use graphics to organize the names of the directors and the films they made.

<h3>Writing code in python:</h3>

<em>import pandas as pd</em>

<em>df = pd.read_csv('all_data/movie_metadata.csv')</em>

<em>print('Shape: ',df.shape)</em>

<em>df.head()</em>

<em>name = df[['director_name','imdb_score']] </em>

<em>directors = name.groupby(['director_name']).mean() </em>

<em>directors = directors.reset_index() </em>

<em>directors = directors.stack()</em>

<em>print(directors,'\n')</em>

<em>print(type(directors),'\n')</em>

<em>print(directors[directors=='Steven Spielberg'],'\n')</em>

<em>print(directors[102]['imdb_score'])</em>

<em>miyazaki = df[df['director_name'] == 'Hayao Miyazaki']['movie_title']</em>

<em>print(miyazaki)</em>

<em>pivot_agg = pd.pivot_table(df,values='imdb_score', index=['country'], columns=['director_name'],aggfunc=np.median)</em>

<em>pivot_agg</em>

<em>gladiator_duration = df[df['movie_title']=='Gladiator\xa0']['duration'] </em>

<em>gladiator_duration</em>

See more about python at brainly.com/question/18502436

#SPJ1

6 0
2 years ago
Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value o
Fantom [35]

Complete question:

Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value of a word is the sum of each tile's points added to any points provided by the word's placement on the game board. Write a program using the given dictionary of letters and point values that takes a word as input and outputs the base total value of the word (before being put onto a board). Ex:  If the input is:  PYTHON

the output is: 14

part of the code:

tile_dict = { 'A': 1, 'B': 3, 'C': 3, 'D': 2, 'E': 1, 'F': 4, 'G': 2, 'H': 4, 'I': 1, 'J': 8,  'K': 5, 'L': 1, 'M': 3, 'N': 1, 'O': 1, 'P': 3, 'Q': 10, 'R': 1, 'S': 1, 'T': 1,  'U': 1, 'V': 4, 'W': 4, 'X': 8, 'Y': 4, 'Z': 10 }

Answer:

Complete the program as thus:

word = input("Word: ").upper()

points = 0

for i in range(len(word)):

   for key, value in tile_dict.items():

       if key == word[i]:

           points+=value

           break

print("Points: "+str(points))

Explanation:

This gets input from the user in capital letters

word = input("Word: ").upper()

This initializes the number of points to 0

points = 0

This iterates through the letters of the input word

for i in range(len(word)):

For every letter, this iterates through the dictionary

   for key, value in tile_dict.items():

This locates each letters

       if key == word[i]:

This adds the point

           points+=value

The inner loop is exited

           break

This prints the total points

print("Points: "+str(points))

6 0
3 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

5 0
3 years ago
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