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Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.

Juliette [100K]

Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

system("Pause");

return 0;

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

if (hour < 13 && hour > 0)

{hour = hour + 12;

return hour;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input hour again in correct 12 hour format: ";

cin >> hour;

invalidHr(hour);

throw 10;

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

if (min < 60 && min > 0)

{return min;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

cin >> min;

invalidMin(min);

throw 20;

return 0;

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

if (sec < 60 && sec > 0)

{return sec;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

cin >> sec;

invalidSec(sec);

throw 30;

return 0;

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0

3 years ago

Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given

Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = $\frac{1}{6}*334.8 = 55.8 ft^3$

Volume of sand = $\frac{2}{6}*334.8 = 111.6 ft^3$

Volume of gravel = $\frac{3}{6}*334.8 = 167.4 ft^3$

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0

3 years ago

There are two machines for sale that you are considering purchasing for your sawmill to produce hardwood flooring. You want to f

devlian [24]

Answer:

Machine 2 has a higher process capability index, it would be best considered for purchase.

Explanation:

Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]

For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7

Cpk= [0.952;0.952]= 0.952

For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3

Cpk= [1.111;3.333]= 1.111

It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.

Since machine 2 has a higher process capability index, it would be best considered for purchase.

4 0

3 years ago

An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side

Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to = 4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × $10^{6}$ + 4 × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × $10^{6}$ - 4 × 10³

lower side frequencies = 3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies - lower side frequencies

total width bandwidth = 8kHz

6 0

3 years ago

A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o

anastassius [24]

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × $10^{-3}$ A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × $10^{-3}$

R2 = 1500Ω

6 0

3 years ago