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Charra [1.4K]
4 years ago
11

(CLO 3—Boolean/Comb. Logic) It is desired to multiplex four different input data lines, a-d, onto one output. Three address line

s, ("x" [MSB] through "z" [LSB]) control input-to-output selection. The three-bit address can be stated as a decimal number ranging from 0 to 7. Input a is MUXed out on address 3, b on address 4, c on 6, d on 7. Draw the MUX circuit below.

Engineering
1 answer:
ANTONII [103]4 years ago
6 0

Answer:

Please find attached file for complete answer solution and explanation of same question.

Explanation:

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Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

                 smallest = values[month];

return smallest;

          }

 

int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl;

  cout << "The smallest amount of rainfall was ";

  cout << getSmallest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl << endl;

  return 0;

}

8 0
4 years ago
A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of
zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6)  = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

So Total net Force on each bolt Fb = P_Eb + Pi

Fb = 4.9 kip + 40 kip

Fb = 44.9 kip

Therefore the net force in every bolt is 44.9 kip

4 0
3 years ago
Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at
MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

4 0
3 years ago
Metal and dirt are not considered contaminants to oll.<br> A) O True<br> B) O False
Likurg_2 [28]

Answer:

true

Explanation:

6 0
3 years ago
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Novay_Z [31]
The monkey is brown!
6 0
3 years ago
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