A cd stores the same number of songs in uncompressed format as in mp3 format.

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

3 years ago

Information is stored on a compact disk in a coded pattern of tiny pits arranged in a track that spirals outward toward the rim

Maksim231197 [3]

Answer:

(i) The rotation speed must stay the same.

(ii) The rotation speed must increase.

(iii) The rotation speed must decrease.

Explanation:

According to Equation

10.10, the angular speed must therefore vary as the laser–lens system moves

radially along the disc. In a typical CD player, the constant speed of the surface at

the point of the laser–lens system is 1.3 m/s.

5 0

2 years ago

Temperature Class Write a Temperature class that will hold a temperature in Fahrenheit and provide methods to get the temperatur

posledela

Answer:

Explanation:

public class Temperature

{

double ftemp;

public int Constructor(double fahrenheit)

{

ftemp = fahrenheit;

return Convert.ToInt32(ftemp);

}

public void setFahrenheit(double fahrenheit)

{

ftemp = fahrenheit;

}

public void getFahrenheit()

{

ftemp = Constructor(ftemp);

}

public void getCelcius()

{

ftemp = (ftemp - 32) * 5 / 9;

}

public void getKelvin()

{

ftemp = (ftemp - 32) * 5 / 9 + 273.15;

}

8 0

2 years ago

Which of the following statements regarding critical paths is true? a. On a specific project, there can be multiple critical pat

iVinArrow [24]

Answer:

Option a

Explanation:

The Critical Path Method is the arrangement of booked exercises that decides the term of the task. These planned exercises must be performed if the venture is to be viewed as a triumph.

Therefore, options b, c, d and e can't be true because:

b. Activities in the Critical Path Method has no or zero slack.

c. The duration of the critical path in CPM determined on the basis of the latest activity and the earliest initiation.

d. The CPM method schedules the activity of the longest duration.

7 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

2 years ago