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3 years ago

Speed limit signs are classified as

Engineering

2 answers:

Ann [662]3 years ago

8 0

The correct answer is B

luda_lava [24]3 years ago

5 0

Answer:

B. Regulatory signs

Explanation:

traffic rules

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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

Explanation:

7 0

3 years ago

Which of the following identifies a criterion needed to become a civil engineer?

denis23 [38]

I’m pretty sure the answer is B

5 0

4 years ago

Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control

NNADVOKAT [17]

Answer:

See explaination

Explanation:

please see attachment for the detailed diagram used in solving the given problem.

It is attached as an attachment.

7 0

3 years ago

Hello it's my new id<br>I am numu

Sonja [21]

Answer:

i am felix

Exp lanation:

nice to meet you

6 0

3 years ago

(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator

MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

$100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\$

Carrier frequency 100kHz

Upper side band

$100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz$

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

$100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz$

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

$100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz$

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

$100k, 99.9k, 99.8k\ \ and \ \99.6kHz$

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

$100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz$

3 0

3 years ago