All categories

3 years ago

In which situation is coauthoring of presentations primarily utilized?

Computers and Technology

2 answers:

FinnZ [79.3K]3 years ago

7 0

Answer:

B) Multiple authors must be able to simultaneously make changes to a presentation.

Explanation:

Co indicates being part of a whole. So this means that two or more authors are working on this project simultaneously. Such as co-management; two bosses working together in an equal position.

Tanya [424]3 years ago

3 0

Answer:

A i belive

Explanation:

You might be interested in

Consider the code fragment below (with nested loops). int sum = 0;for (int i = 1; i < 5; i++) for (int j = 1; j <= i; j++)

sammy [17]

Answer:

Option d is the correct answer for the above question.

Explanation:

The first loop of the program has a second loop and then the statement. In this scenario, the second loop executes for the value of the first loop and the statement executes for the value of the second loop.
The first loop executes 4 times, Then the second loop or inner loop executes n times for the n iteration of the first loop, for example, 1 time for the first iteration of the first loop, 2 times for the second iteration of the first loop and so on.
Then the inner loop executes (1+2+3+4) iteration which gives the result 10 iterations.
The sum initial value is 0 and the "sum++", increase the value of the sum by 1.
So the value of the sum becomes 10 after completing 10 iterations of the inner for loop.
Hence the 10 will be the output. So the Option d is the correct answer while the other is not.

3 0

3 years ago

Convert the following as indicated : (120)10 = (?)2

Nookie1986 [14]

Answer:

1111000

Explanation:

perform the following. write from right to left:

- is the number even? then write down a 0

- is the number odd? then write down a 1 and subtract 1

- divide by 2

- repeat until you reach 0.

So for 120:

120 is even, so write down a 0 and continue with 120/2=60

60 is even, so write down a 0 and continue with 60/2=30

30 is even, so write down a 0 and continue with 30/2=15

15 is odd, so write down a 1 and continue with 14/2=7

7 is odd, so write down a 1 and continue with 6/2=3

3 is odd, so write down a 1 and continue with 2/2=1

1 is odd, so write down a 1 and finish with 0

4 0

3 years ago

These are templates or patterns that make it easier for users to enter data

kupik [55]

Answer:

A piece of code

Explanation:

8 0

3 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

3 years ago

By default the Windows desktop displays programs, task bar and gadgets. T/F

pychu [463]

True, the Windows desktop displays task manager, all programs, computer status, documents...

3 0

3 years ago