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Levart [38]
2 years ago
13

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

limit of 400 parts per million (ppm) in public play areas designed for children. Mike is an inspector, and he takes 50 randomly selected soil samples from a site where they are considering building a playground.
These data show a sample mean of x = 390.25 ppm and standard deviation of Sx = 30.5 ppm. Answer the following:

1. Determine the critical t value and along with the confidence interval for 95% level of

confidence. 2. What does this interval suggest?
Engineering
1 answer:
DENIUS [597]2 years ago
3 0

Answer:

See below

Explanation:

<u>Check One-Sample T-Interval Conditions</u>

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

<u>One-Sample T-Interval Information</u>

  • Formula --> CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})
  • Sample Mean --> \bar{x}=390.25
  • Critical Value --> t^*=2.0096 (given df=n-1=50-1=49 degrees of freedom at a 95% confidence level)
  • Sample Size --> n=50
  • Sample Standard Deviation --> S_x=30.5

<u>Problem 1</u>

The critical t-value, as mentioned previously, would be t^*=2.0096, making the 95% confidence interval equal to CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

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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)

{

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return 1;

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else if (p == 0)

{

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 printf("\n\tEnter meaaage:"):

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 // 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];

 }

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 // 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");

 }

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   close(fd3[1]);

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

   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");

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

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   if(p<0)

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    printf("Chiile_Procrss_2 not cheated");

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   else

   {

     

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

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

    close(fd3[0]);

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

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   }

  }

 }

 close(fd2[1]);

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Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number (Re) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

Re=\frac{\rho v D}{\mu}

where \rhois the density of the fluid, \mu is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm

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