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slavikrds [6]
3 years ago
14

Communication is the transmission of messages to large audiences.

Computers and Technology
1 answer:
masha68 [24]3 years ago
3 0
Communication doesn't necessarily mean you're communicating to a large audience. So that is false.
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Which single OSPF network statement will correctly include all the interfaces on a device whose IP addresses only begin with a 1
tester [92]

Answer:

network 10.10.8.0 0.0.3.255 area 0

this will include all the interfaces on a device whose IP addresses only begin with a 10.10.8, 10.10.9, 10.10.10, or 10.10.11.

Explanation:

<em>show ip ospf interface </em>

<em>show ip ospf interface brief</em>

these commands are used to display the interfaces that have been enabled into local ospf . it also shows explanation about them by brief command mentioned above.

<em />

4 0
3 years ago
Minor changes to objects which only take effect at render time are called:
Ad libitum [116K]

Answer:

modifiers

Explanation:

The modifiers are the change that is minor with respect to the objects that have an impact at the render time

Therefore as per the given situation, the last option is correct as it fits to the current scenario

And, the rest of the options would be incorrect

The same is to be considered and relevant

Hence, the modifier is the correct option

7 0
3 years ago
Which of these statements would create a variable but not assign anything to it?
Ugo [173]

Answer:

variable = null

Explanation:

that one would fasho

7 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
Write a C class, Flower, that has three member variables of type string, int, and float, which respectively represent the name o
Ulleksa [173]

Answer and Explanation:

C is a low level language and does not have classes. The language isn't based on object oriented programming(OOP) but is actually a foundation for higher level languages that adopt OOP like c++. Using c++ programming language we can implement the class, flower, with its three variables/properties and functions/methods since it is an object oriented programming language.

3 0
3 years ago
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