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3 years ago

What cell phone technology is the most popular in the united states?

2 answers:

Nookie1986 [14]3 years ago

5 0

Like all the zero dollar phones and a 2.0 MP camera or lower are the most common every where in north america.

phone like the Blackberry, iPhone, and maybe HTC are in a different class (smartphones)
mostly they are going to be flips and sliders, because they look cooler. the full keyboard sliders are garbing the young teenage girls and some guys with it.

my previous phones:
Sony Ericsson W580i
Sony Ericsson G900
Current: HTC Dream
i want the Sony Ericsson idou, or the X2 Cheers,Irys ;3

Alchen [17]3 years ago

3 0

IPhone and Android. ....

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Can someone help me with this pls

Dimas [21]

I know the first one shown in the picture but I can't see the one under it I can help you if you can show the whole problem under the first problem.

5 0

3 years ago

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Letter Frequency Write a function that will take a string and return a count of each letter in the string. For example, "my dog

Nostrana [21]

Answer:

Check the explanation

Explanation:

Code to copy:

// ConsoleApplication7.cpp : Defines the entry point for the console application.

#include "stdafx.h"

#include <iostream>

#include <cstring>

#include <string>

using namespace std;

int * letterFrequency(char s[]);

int main()

{

int *freq_letters;

char *s = new char[100];

freq_letters = new int[26];

//To read the input string in order to find the frequency of each letter

cout << "Enter a string: ";

cin.getline(s, ' ');

//call the function to find the occurrence of each alphabet

freq_letters = letterFrequency(s);

//Display the count

cout << "Letter Frequency " << endl;

for (int i = 0; i < 26; i++)

{

//Constriant to check if the letter appeared at least once in the string and so printing the frequency of its occurence

if (freq_letters[i] != 0)

cout << " " << static_cast<char>(i + 'a') << " " << freq_letters[i] << endl;

}

system("pause");

return 0;

}

//Define the function to find occurrence of each letter in the input string

int * letterFrequency(char s[])

{

int *occurrence_array;

//to store the output of occurrences for each alphabet

occurrence_array = new int[26];

// to store the count of occurrence for each letter temporarily

int letter_count;

// for loop to check the occurrence for all 26 alphabets

for (int i = 0; i < 26; i++)

{

letter_count = 0;

for (int j = 0; j < strlen(s); j++)

{

/*comparing the ascii values of each alphabet with every character from the string by converting it to lower case i.e. case insensitive*/

if (int('a') + i == int(tolower(s[j])))

letter_count++;

}

occurrence_array[i] = letter_count;//To store the count calculated for each alphabet

}

return occurrence_array;

}

The following below code screenshot and output shows that when the string is entered, the output will shows each lettercount irrespective of whether the letter is in capitals or small and does not count non-letter characters (i.e spaces, punctuations etc.)

6 0

4 years ago

Sarah maintains a blog about her soap-making business, and she has hired someone to create a database for this business. She mak

ivann1987 [24]

The database planner would most likely create a table that contains customer contact information since these would be the individuals who placed an order for Sarah's products after visiting her blog.

4 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

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

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 $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
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

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

The resulting run has length 6000 records.

Part b

For step 1

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

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

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

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

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.

For step 2

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

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

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

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

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)

Now the total time for merging is given as

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

Total time in merging is 249 seconds seconds

5 0

3 years ago

What is the purpose of the Zoom dialog box? to put the selected range in ascending order in a query to put the selected range in

AlekseyPX

The zoom dialog box helps to enter text and expressions in tight places, such as a property sheet or Query Design view. You can resize the zoom box and change the font. And the zoom box remembers the size and font each time you use it.

Use the

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3 years ago

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