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worty [1.4K]
3 years ago
9

What is the most efficient way to include a space after each paragraph

Computers and Technology
1 answer:
3241004551 [841]3 years ago
4 0

To skip a line, then write the rest like this!


See? Good spacing, right?

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You are hired to train a group of new users to become technicians. One of your lessons is on how to construct Cat5 and Cat6 Ethe
Greeley [361]

Answer:

To ensure that the trainees are well acquainted with how to differentiate 568A cables and 568B cables, you must ensure they know appropriate color scheme pattern for the both cables. Also, they are expected to know the correct terminology when using both cables, as 568A is known as a crossover cable and 568B is known as a straight-through cable.

Explanation:

3 0
3 years ago
Write a method called findNames that meets the following specs: It takes two arguments: a list of strings, allNames, and a strin
Anna11 [10]

Answer:

public class Solution {

   public static void main(String args[]) {

       String[] allNames = new String[]{"Bob Smith", "Elroy Jetson", "Christina Johnson", "Rachael Baker", "cHRis", "Chris Conly"};

       String searchString = "cHRis";

       

       findNames(allNames, searchString);

   }

   

   public static void findNames(String[] listOfName, String nameToFind){

       ArrayList<String> resultName = new ArrayList<String>();

       

       for(String name : listOfName){

           if (name.toLowerCase().contains(nameToFind.toLowerCase())){

               resultName.add(name);

           }

       }

       

       for(String result : resultName){

           System.out.println(result);

       }

   }

}

Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

The method findNames is defined and it accept two parameters, an array containing list of names and the name to search for.

Inside the findNames method, we initialized and assigned an ArrayList called resultName. The resultName variable is to hold list of element found that contain the searchString.

The first for-loop goes through the elements in the allNames array and compare it with the searchString. If any element is found containing the searchString; it is added to the resultName variable.

The second for-loop goes through the elements of the resultName array and output it. The output is empty if no element was found added to the resultName variable.

During comparison, the both string were converted to lower case before the comparison because same lowercase character does not equal same uppercase character. For instance 'A' is not same as 'a'.

8 0
3 years ago
When a slide is inserted, it appears A. at the end of the presentation. B. before the selected slide. C. after the selected slid
padilas [110]
C. after the selected slide
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4 years ago
Read 2 more answers
Why do some people have random numbers as their usernames?
Angelina_Jolie [31]
Mostly for privacy or to not get caught looking up answers probably
3 0
3 years ago
Read 2 more answers
A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an
yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

                                       CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

                    = 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x  10⁹ x 0.75 / 3 Ghz

                  = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

     Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

                = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that  CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore,   numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x  10⁹ / 0.75  = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall  that:

MIPS = Clock rate / CPI X 10⁶

  So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

        MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

  We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

                  MFLOPS₁ = 1777.6

                  MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0
3 years ago
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