Please help with these linear equations

PLEASE HELP QUICK WILL MARK BRAINLIEST.

ioda

Answer:

4.57

Step-by-step explanation:

7 can go into 32 4 times but your remainder is going to be 5

5 0

3 years ago

Which of the following best describes the perpendicular lines?

AysviL [449]

Answer:

A perpendicular line is a line that meets a 90 degree angle.

Step-by-step explanation:

Think of the letter L

the top stick of the letter is perpendicular to the bottom stick and it makes a 90 degree angle.

Thats the best way I can explain it.

Hope this helps and please mark me brainliest if it did :)

8 0

3 years ago

Read 2 more answers

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a

REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

$p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157$

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

$p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105$

Distribution of the difference:

$p = p_T - p_A = 0.73 - 0.47 = 0.26$

$s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019$

Confidence interval:

Is given by:

$p \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$p - 1.96s = 0.26 - 1.96*0.019 = 0.223$

Upper bound:

$p + 1.96s = 0.26 + 1.96*0.019 = 0.297$

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0

3 years ago

Research about 5 monuments around the world where Roman numerals are used to depict the year in which they were built explain in

posledela

Answer:

Modern uses. Roman numerals are still used today and can be found in many places. They are still used in almost all cases for the copyright date on films, television programmes, and videos - for example MCMLXXXVI for 1986.

4 0

3 years ago

a. For a file whose size (in terms of number of bytes) is less than or equal to 12, then duplicate the contents of the file so t

pickupchik [31]

Answer:

#include <stdio.h>

int main()

{

int check;

char* dirname = "Subdir1";

check = mkdir(dirname,0766);

// check if directory is created or not

if (!check)

{

printf("Subdir1 Directory created\n");

char *dirname1="/Subdir1/D1";

if(mkdir("/Subdir1/D1", 0777)==-1)

{

printf("D1 Directory created\n");

int A1 = open ("/Subdir1/D1/A1.txt", 0666);

write(A1, "abcdefghij\n", strlen("abcdefghij\n"));

close(A1);

int A2 = open ("/Subdir1/D1/A2.txt", 0666);

write(A2, "you are beautiful\n", strlen("you are beautiful\n"));

close(A2);

int A3 = open ("/Subdir1/D1/A3.txt", 0666);

write(A3, "this is terrible!\n", strlen("this is terrible!\n"));

close(A3);

}

else

{

printf("Unable to create directory D1\n");

exit(1);

}

if(mkdir("/Subdir1/D2", 0777)==-1)

{

printf("D2 Directory created\n");

int B1 = open ("/Subdir1/D2/B1.ccc", 0664);

write(B1, "abcdefghij\n", strlen("abcdefghij\n"));

close(B1);

int B2 = open ("/Subdir1/D2/B2.ccc", 0664);

write(B2, "you are beautiful\n", strlen("you are beautiful\n"));

close(B2);

int B3 = open ("/Subdir1/D2/B3.ccc", 0664);

write(B3, "this is terrible!\n", strlen("this is terrible!\n"));

close(B3);

}

else

{

printf("Unable to create directory D2\n");

exit(1);

}

if(mkdir("/Subdir1/D3", 0777)==-1)

{

printf("D3 Directory created\n");

int C1 = open ("/Subdir1/D3/C1.txt", 0644);

write(C1, "abcdefghij\n", strlen("abcdefghij\n"));

close(C1);

int C2 = open ("/Subdir1/D3/C2.ddd", 0644);

write(C2, "you are beautiful\n", strlen("you are beautiful\n"));

close(C2);

int C3 = open ("/Subdir1/D3/C3.txt", 0644);

write(C3, "this is terrible!\n", strlen("this is terrible!\n"));

close(C3);

}

else

{

printf("Unable to create directory D2\n");

exit(1);

}

else {

printf("Unable to create directory Subdir1\n");

exit(1);

}

return 0;

}

Program 2:

#include <stdio.h>

long int findSize(char file_name[])

{

// opening the file in read mode

FILE* fp = fopen(file_name, "r");

// checking if the file exist or not

if (fp == NULL) {

printf("File Not Found!\n");

return -1;

}

fseek(fp, 0L, SEEK_END);

// calculating the size of the file

long int res = ftell(fp);

// closing the file

fclose(fp);

return res;

}

// Driver code

int main()

{

FILE *fptr1;

char c;

char file_name[] = {"D1.txt"};

long int res = findSize(file_name);

if (res != -1)

{

printf("Size of the file is %ld bytes \n", res);

if(res<=12)

{

fptr1 = fopen("D1.txt", "r");

c = fgetc(fptr1);

while (c != EOF)

{

fputc(c, fptr1);

c = fgetc(fptr1);

printf(c);

}

return 0;

}

4 0

3 years ago