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MAVERICK [17]
1 year ago
14

A(n) ____ is an electronic device, operating under the control of instructions stored in its own memory, that can accept data, p

rocess the data according to specified rules, produce results, and store the results for future use.
Computers and Technology
1 answer:
AveGali [126]1 year ago
8 0
A computer. 
Core components of a computer
Processor (cpu): can compute instructions 
RAM (random access memory): Holds running processes. 
HDD( hard disk drive): Stores data, e.g. pictures, documents, games etc. 
Motherboard: Allows devices to communicate with each other via a "bus". 
Power Supply (PSU): converts AC current to DC, providing the computer with power. 
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How did josh norman and mike keller provide coverage of katrina?
Digiron [165]
<span>They wrote live updates to a blog, so D. The blog was called "Eye of the Storm." It was written as a series of personal musings and photographs of places where the storm hit- first hand accounts of the disaster. They received a journalism award for it.</span>
6 0
2 years ago
Read 2 more answers
A file named data.txt contains an unknown number of lines, each consisting of a single integer. Write some code that creates two
Damm [24]

Answer:

I am doing it with python.

Explanation:

nums = '9 -8 -7 -6 -5 -4 -2 0 1 5 9 6 7 4'

myfile = open('data.txt', 'w')

myfile.write(nums)

myfile.close()

myfile = open('data.txt', 'r')

num1 = (myfile.read())

num1 = num1.split()

print(num1)

print(type(num1))

for x in num1:

   x = int(x)

   if x < 0:

       minus = open('dataminus.txt', 'a')

       minus.write(str(x) + ' ')

       minus.close()

   elif x>= 0:

       plus = open('dataplus.txt', 'a')

       plus.write(str(x)+' ')

       plus.close()

3 0
2 years ago
Read 2 more answers
Question _.1 with 1 blankMiguel y Maru están muy cansados. Question 2 with 1 blankFelipe es muy joven. Question 3 with 1 blankJi
Alekssandra [29.7K]

Answer:

Miguel y Maru están muy cansados.  - Miguel y Maru están cansadísimos

Felipe es muy joven. - Felipe es jovencísimo

Jimena es muy inteligente. - Jimena es inteligentísima

La madre de Marissa está muy contenta. - La madre de Marissa está contentísima

Estoy muy aburrido. - Estoy aburridísimo

Explanation:

In this activity we have to switch the statements to the absolute superlative of the expressions. In Spanish we can add the suffix -ísimo to an adjective to refer to the highest degree of something. It can be translated in ENglish to "really, extremely, super or quie". The statements in English are:

- Miguel and Maru are very tired - Miguel and Mary are extremely tired

- Felipe is very young - Felipe is super young

- Jimena is very smart - Jimena is really smart

- Marissa´s mother is very happy - Marissa´s mother is extremely happy

- I´m very bored - I´m super bored

8 0
2 years ago
Write a public static method named printArray, that takes two arguments. The first argument is an Array of int and the second ar
Serjik [45]

Answer:

Written in Java

public static void printArray(int myarr[], String s){

       for(int i = 0; i<myarr.length;i++){

           System.out.print(myarr[i]+s);

       }

   }

Explanation:

This defines the static method alongside the array and the string variable

public static void printArray(int myarr[], String s){

The following iteration iterates through the elements of the array

       for(int i = 0; i<myarr.length;i++){

This line prints each element of the array followed by the string literal

           System.out.print(myarr[i]+s);

       }

   }

The method can be called from main using:

<em>printArray(myarr,s);</em>

Where myarr and s are local variables of the main

5 0
2 years ago
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
1 year ago
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