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3 years ago

A business owned by one people is a

Engineering

2 answers:

Whitepunk [10]3 years ago

6 0

Answer:

A sole proprietorship, also known as the sole trader, individual entrepreneurship or proprietorship, is a type of enterprise that is owned and run by one person and in which there is no legal distinction between the owner and the business entity.

Explanation:

Damm [24]3 years ago

3 0

Answer:

A business owned by one people is a-Sole proprietorship

Explanation:

This is correct i did it for my electives class which was career reaserch and decision making.

brainliest?

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Yall know what this is called?

aliya0001 [1]

Answer:

oof no bro

Explanation:

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3 years ago

if you had 100 B size sheets and you cut them into A size sheets, how many sheets of A size paper would you have

castortr0y [4]

Answer:

200

Explanation:

A size sheets (also known as letter size) are 8.5 inches by 11 inches.

B size sheets (also known as ledger size) are 11 inches by 17 inches.

One B size sheet is twice as large as a A size sheet. So if you have 100 B size sheets and cut each one in half, you'll get 200 A size sheets.

8 0

3 years ago

Need help with both giving out brainlest for the people to help me

sammy [17]

The first one is d or the 4th answer choice and the second one is false. Hope this helps!

4 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

4 years ago

CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic

valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd2)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd3)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd4)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

p = fork();

if (p < 0)

{

fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

close(fd1[0]);// Close reading end of first pipe

char concat_str[100];

printf("\n\tEnter meaaage:"):

scanf("%s",concat_str);

write(fd1[1], concat_str, strlen(concat_str)+1);

// Concatenate a fixed string with it

int k = strlen(concat_str);

int i;

for (i=0; i<strlen(fixed_str); i++)

{

concat_str[k++] = fixed_str[i];

}

concat_str[k] = '\0';//string ends with '\0'

// Close both writting ends

close(fd1[1]);

wait(NULL);

//.......................................................................

close(fd2[1]);

read(fd2[0], concat_str, 100);

if(strcmp(concat_str,"invalid")==0)

{

printf("\n\tmessage not send");

}

else

{

printf("\n\tmessage send to prog_2(child_2).");

}

close(fd2[0]);//close reading end of pipe 2

exit(0);

}

else

{

close(fd1[1]);//Close writting end of first pipe

char concat_str[100];

read(fd1[0], concal_str, strlen(concat_str)+1);

close(fd1[0]);

close(fd2[0]);//Close writing end of second pipe

if(/*check if msg is valid or not*/)

{

//if not then

write(fd2[1], "invalid",sizeof(concat_str));

return 0;

}

else

{

//if yes then

write(fd2[1], "valid",sizeof(concat_str));

close(fd2[1]);

q=fork();//create chile process 2

if(q>0)

{

close(fd3[0]);/*close read head offd3[] */

write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

close(fd3[1]);

wait(NULL);//wait till child_process_2 send ACK

//...........................................................

close(fd4[1]);

read(fd4[0],concat_str,100);

close(fd4[0]);

if(sctcmp(concat_str,"ack")==0)

{

printf("Messageof child process_1 is received by child process_2");

}

else

{

printf("Messageof child process_1 is not received by child process_2");

}

else

{

if(p<0)

{

printf("Chiile_Procrss_2 not cheated");

}

else

{

close(fd3[1]);//Close writing end of first pipe

char concat_str[100];

read(fd3[0], concal_str, strlen(concat_str)+1);

close(fd3[0]);

close(fd4[0]);//Close writing end of second pipe

write(fd4[1], "ack",sizeof(concat_str));

}

close(fd2[1]);

}

8 0

4 years ago