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DiKsa [7]
3 years ago
7

Choose the correct line of code.

Computers and Technology
1 answer:
Naddika [18.5K]3 years ago
5 0
Def_pow_(self,b): if not then I’m sorry
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The posture is how someone sits at his or her computer.

Hope this helps ;-))

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A check list should be based on that apply to your industry.
frozen [14]
Can you please elaborate? This is as your question isn't very clear.
6 0
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Write a program in c++ to displaypascal’s triangle?
Harman [31]

<u> C++ Program to Print Pascal's Triangle</u>

 #include<iostream> //header file

using namespace std;

//driver function  

int main()

{

   int r;/*declaring r for Number of rows*/

   cout << "Enter the number of rows : ";

   cin >> r;

   cout << endl;

 

   for (int a = 0; a < r; a++)

   {

       int value = 1;

       for (int b = 1; b < (r - a); b++) /*Printing the indentation space*/

       {

           cout << "   ";

       }

       for (int c = 0; c <= a; c++) /*Finding value of binomial coefficient*/

       {

           cout << "      " << value;

           value = value * (a - c) / (c + 1);

       }

       cout << endl << endl;

   }

   cout << endl;

   return 0;

}

<u>Output</u>

<u>Enter the number of rows :  5</u>

                 1

              1      1

           1      2      1

        1      3      3      1

     1      4      6      4      1

7 0
3 years ago
Which of the following is true of a bit in data storage?
Trava [24]

Answer:

You didn't give any choice as it looks a multiple choice question

6 0
3 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
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