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spayn [35]
3 years ago
12

The concepts of culturally approved goals and the legitimate means of achieving these goals was devised by

Computers and Technology
2 answers:
xxMikexx [17]3 years ago
6 0

the answer is not C the anwser i have no idea what it is but on my test i got it wrong.

svetlana [45]3 years ago
5 0

The answer to your question is letter B.

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In c please
Paraphin [41]

Answer:

#include <stdio.h>

#include <ctype.h>

void printHistogram(int counters[]) {

   int largest = 0;

   int row,i;

   for (i = 0; i < 26; i++) {

       if (counters[i] > largest) {

           largest = counters[i];

       }

   }

   for (row = largest; row > 0; row--) {

       for (i = 0; i < 26; i++) {

           if (counters[i] >= row) {

               putchar(254);

           }

           else {

               putchar(32);

           }

           putchar(32);

       }

       putchar('\n');

   }

   for (i = 0; i < 26; i++) {

       putchar('a' + i);

       putchar(32);

   }

}

int main() {

   int counters[26] = { 0 };

   int i;

   char c;

   FILE* f;

   fopen_s(&f, "story.txt", "r");

   while (!feof(f)) {

       c = tolower(fgetc(f));

       if (c >= 'a' && c <= 'z') {

           counters[c-'a']++;

       }

   }

   for (i = 0; i < 26; i++) {

       printf("%c was used %d times.\n", 'a'+i, counters[i]);

   }

   printf("\nHere is a histogram:\n");

   printHistogram(counters);

}

5 0
2 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
Proponents of Internet freedom see its _____________ as providing protection for unpopular expression; proponents of greater Int
diamong [38]

Answer:

anonymity

Explanation:

<h2><u>Fill in  the blanks </u></h2>

Proponents of Internet freedom see its <u>anonymity </u>as providing protection for unpopular expression; proponents of greater Internet control see it as the Internet's greatest danger.

6 0
3 years ago
Help me find the right answer please.
timama [110]

Answer:

I'm not sure

Explanation:

I'm good at math but I do not see any numbers to work with. SORRY

5 0
3 years ago
A service specialist from your company calls you from a customer's site. He is attempting to install an upgrade to the software,
IgorLugansk [536]

Answer:

kill 1000

Explanation:

3 0
3 years ago
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