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3 years ago

"The number of hardware chips needed for multiple digit display can be minimized by using the technique called ________"

Computers and Technology

1 answer:

Dafna1 [17]3 years ago

4 0

Answer: Multiplexing

Explanation:

By using the multiplexing technique we can easily use the multiple digit display in the input pins of the hardware. We can use this technique to display the several digits and the extra hardware is require which is proportional to the number of several digit to be display.

As, hardware use only limited number of input and output pins. Therefore, multiplexing is the technique used the number of hardware chips need for multiple digit display and can be minimize using this technique.

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3 years ago

Read 2 more answers

a. displays the sum of all even numbers between 2 and 100 (inclusive). b. displays the sum of all squares between 1 and 100 (inc

LuckyWell [14K]

Answer:

The program required is in the explanation segment.

Explanation:

Program :

import math

# a. displays the sum of all even numbers between 2 and 100 (inclusive).

print("All even numbers from 2 to 100 inclusive ")

sum=0

i=2

while i<=100:

if i %2 ==0:

sum=sum+i

print(i,end=" ")

i=i+1

print("\nThe sum of all even numbers between 2 and 100 (inclusive) :",sum);

#b. displays the sum of all squares between 1 and 100 (inclusive).

print("\nAll squares numbers from 1 to 100 inclusive:")

i=1

sum=0

while i<=100:

print(i*i,end=" ")

i=i+1

sum=sum+(i*i)

print("\n\nThe sum of all squares between 1 and 100 (inclusive) is :",sum)

#c. displays the powers of 2 from 1 up to 256.

print("\nAll powers of 2 from 2 ** 0 to 2 ** 8:")

i=0

while True:

p=math.pow(2,i)

if p>256:

break

print("2 ** ",i," is ",int(p))

i=i+1

#d. displays the sum of all odd numbers between a and b (inclusive), where a and b are inputs

print("\nCompute the sum of all odd integers between two intgers ")

a=int(input("Enter an integer:"))

b=int(input("Enter another integer: "))

count = 0

temp=a

sum=0

while a<=b:

if a%2!=0:

print(a,end=" ")

sum=sum+a

a=a+1

print("\nThe total of the odd numbers from ", temp ," to ", b ,"is",sum)

#e.displays the sum of all odd digits of an input. (For example, if the input is 32677, the sum would be 3 + 7 + 7 = 17.)

print("\nCompute the sum of the odd digits in an integer ")

n=int(input("Enter an integer:"))

count=0

sum=0

temp=n

while n!=0:

rem = n%10

if rem%2!=0:

sum=sum+rem

count=count+1

n=int(n/10)

print("Sum of the odd digits is ",sum)

print("The total of the odd digits in ",temp," is ",count)

4 0

3 years ago

Flesh out the body of the print_seconds function so that it prints the total amount of seconds given the hours, minutes, and sec

fiasKO [112]

Answer:

Step by step explanation along with code and output is provided below

Explanation:

#include<iostream>

using namespace std;

// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three

void print_seconds(int hours, int mints, int seconds)

{

int total_seconds= hours*3600 + mints*60 + seconds;

cout<<"Total seconds are: "<<total_seconds<<endl;

}

// test code

// user inputs hours, minutes and seconds and can also leave any of them by entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.

int main()

{

int h,m,s;

cout<<"enter hours if any or enter 0"<<endl;

cin>>h;

cout<<"enter mints if any or enter 0"<<endl;

cin>>m;

cout<<"enter seconds if any or enter 0"<<endl;

cin>>s;

print_seconds(h,m,s);

return 0;

}

Output:

enter hours if any or enter 0

enter mints if any or enter 0

enter seconds if any or enter 0

Total seconds are: 8710

8 0

3 years ago