What electronic appliances at your home / school can be controlled remotely? Name any 4

Implement one array of the English alphabet (26 characters). a) Create a function, getLetters() to cast and generate the array.

WARRIOR [948]

#include <fstream>

#include <iostream>

#include <iomanip>

#include <cstring>

#include <cstdlib>

using namespace std;

// Function Declarations

void display(char alphabets[],int MAX_SIZE);

void reverse(char alphabets[],int MAX_SIZE);

void swap(char &ch1,char &ch2);

int main() {

//Declaring constant

const int MAX_SIZE=26;

//Declaring a char array

char alphabets[MAX_SIZE];

//Populating the array with alphabets

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

{

alphabets[i]=(char)(65+i);

}

cout<<"Original: ";

display(alphabets,MAX_SIZE);

reverse(alphabets,MAX_SIZE);

cout<<"Reversed: ";

display(alphabets,MAX_SIZE);

return 0;

}

//This function will display the array contents

void display(char alphabets[],int MAX_SIZE)

{

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

{

cout<<alphabets[i]<<" ";

}

cout<<endl;

}

//This function will reverse the array elements

void reverse(char alphabets[],int MAX_SIZE)

{

int first,last;

first=0;

last=MAX_SIZE-1;

while(first<last)

{

swap(alphabets[first],alphabets[last]);

first++;

last--;

}

void swap(char &ch1,char &ch2)

{

char temp;

temp=ch1;

ch1=ch2;

ch2=temp;

6 0

1 year ago

Não basta tornar-se móvel hoje em dia. Ao se organizar para a implantação de tecnologias no contexto da mobilidade, é preciso fi

Ad libitum [116K]

Answer:

Alternativa 2:

4 - 1 - 3 - 2.

Explanation:

4 0

3 years ago

Consider the latency model for static congestion windows. If the server receives an acknowledgement for the first data segment i

Angelina_Jolie [31]

Answer:

Latency of an object O is shown below.

Explanation:

W segment and stalled state transmits nothing and waits for acknowledgement. The latency is 2 R TT + the time required for server that are using to transmit the object + the amount of time when server is in stalled state. Let K be the number of window that is K= O/WS .The serveries stalled state where K-1 is period of ime with period lasting RTT-(W-1)S/R

latency = 2RTT +O/R +(K-1)[S/R +RTT - WS/R]

After combining the two case

latency = 2 RTT + O/R + (K-1)[S/R +RTT - WS/R]

where [x] means maximum of (x.0). This is the complete ananlysis of the static windows.

server time for transmit the object is (K-1)[S/R +RTT - WS/R]

7 0

3 years ago

A testing lab wishes to test two experimental brans of outdoor pain long each wiil last befor fading . The testing lab makes six

Simora [160]

Answer:

The answer is " $\bold{Brand \ A \ (35, 350, 18.7) \ \ Brand \ B \ (35, 50, 7.07)}$ "

Explanation:

Calculating the mean for brand A:

$\to \bar{X_{A}}=\frac{10+60+50+30+40+20}{6} =\frac{210}{6}=35$

Calculating the Variance for brand A:

$\sigma_{A}^{2}=\frac{\left ( 10-35 \right )^{2}+\left ( 60-35 \right )^{2}+\left ( 50-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 20-35 \right )^{2}}{5} \\\\$

$=\frac{\left ( -25 \right )^{2}+\left ( 25 \right )^{2}+\left ( 15\right )^{2}+\left ( -5 \right )^{2}+\left ( 5 \right )^{2}+\left ( 15 \right )^{2}}{5} \\\\ =\frac{625+ 625+225+25+25+225}{5} \\\\ =\frac{1750}{50}\\\\=350$

Calculating the Standard deviation:

$\sigma _{A}=\sqrt{\sigma _{A}^{2}}=18.7$

Calculating the Mean for brand B:

$\bar{X_{B}}=\frac{35+45+30+35+40+25}{6}=\frac{210}{6}=35$

Calculating the Variance for brand B:

$\sigma_{B} ^{2}=\frac{\left ( 35-35 \right )^{2}+\left ( 45-35 \right )^{2}+\left ( 30-35 \right )^{2}+\left ( 35-35 \right )^{2}+\left ( 40-35 \right )^{2}+\left ( 25-35 \right )^{2}}{5}$

$=\frac{\left ( 0 \right )^{2}+\left ( 10 \right )^{2}+\left ( -5 \right )^{2}+\left (0 \right )^{2}+\left ( 5 \right )^{2}+\left ( -10 \right )^{2}}{5}\\\\=\frac{0+100+25+0+25+100}{5}\\\\=\frac{100+25+25+100}{5}\\\\=\frac{250}{5}\\\\=50$

Calculating the Standard deviation:

$\sigma _{B}=\sqrt{\sigma _{B}^{2}}=7.07$

4 0

3 years ago

Design a 16K x 8 memory using the following ICs:

Montano1993 [528]

Answer:

Required memory size is 16k x 8

16k = 24 x 210 = 214

Hence, No. of address lines = 14

No. of data lines = 8

a) Size of IC 1024 x 1

Total number of ICs required = 16k x 8 / 1024 x 1 = 16 x 8 = 128

b) Size of IC 2k x 4

Total number of ICs required = 16k x 8 / 2k x 4 = 8 x 2 = 16

c) Size of IC 1k x 8

Total number of ICs required = 16k x 8 / 1k x 8 = 16 x 1 = 16

Explanation:

For a, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1024 memory locations present in a IC.

For b, 11 address lines from A0 to A10 are used to select any one of the memory location out of 2k=2048 memory locations present in a IC.

For c, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1k=1024 memory locations present in a IC.

3 0

3 years ago