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4 years ago

Write an ALP to separate odd and even numbers from an array of N numbers; arrange odd

Engineering

1 answer:

Marta_Voda [28]4 years ago

6 0

Below is the program to separate odd and even numbers

Explanation:

L1:

mov ah,00

mov al,[BX]

mov dl,al

div dh

cmp ah,00

je EVEN1

mov [DI],dl

add OddAdd,dl

INC DI

INC BX

Loop L1

jmp CAL

EVEN1:

mov [SI],dl

add Even Add,dl

INC SI

INC BX

Loop L1

CAL:

mov ax,0000

mov bx,0000

mov al,OddAdd

mov bl,EvenAdd

MOV ax,4C00h

int 21h

end

The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.

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True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'

defon

Answer:

true

Explanation:

8 0

3 years ago

True or false 19. Closed systems rely on feedback from outside of the system to operate.

Furkat [3]

Answer: True

Explanation: Closed loop relies on feedback from PNS to make modifications in the movement, open loop allows action in the absence of feedback, 2. ... Closed loop can change the initial commands, open loop can not change the initial commands.

6 0

3 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67

defon

Answer:

#include <iostream>

#include <string>

using namespace std;

bool isPalindrome(string str)

{

int length = str.length();

for (int i = 0; i < length / 2; i++)

{

if (tolower(str[i]) != tolower(str[length - 1 - i]))

return false;

}

return true;

}

int main()

{

string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};

int i;

for(i=0; i<6; i++)

{

//Testing function

if(isPalindrome(s[i]))

{

cout << "\n " << s[i] << " is a palindrome... \n";

}

else

{

cout << "\n " << s[i] << " is not a palindrome... \n";

}

return 0;

}

5 0

3 years ago

Does somebody know how to do this?

maksim [4K]

No I don’t sorry, I hope you do well

4 0

3 years ago