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2 years ago

How can TestOut's Microsoft Excel course help me become more proficient?

Computers and Technology

1 answer:

nata0808 [166]2 years ago

8 0

Answer:

Explanation:

di ko alam eh HAHAHAHA

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What is the output of this program? Assume the user enters 3, 6, and 11.

erica [24]

Answer:

14.0

Explanation:

3 0

1 year ago

How to change high school to primary school

poizon [28]

Answer:

If you are using Desktop then go to your profile and press the button that says edit profile. And then you will see button that says level.

The levels are

Middle School
Highschool &
College

Hope this helps❤❤❤

6 0

3 years ago

What set of code correctly initializes all elements of the array ar to the value 0, given the declaration

Veronika [31]

Answer:

The correct answer is:

C. ndx = 0;

while (ndx < 3) {

ar[ndx] = 0;

ndx++;

}

Explanation:

The declaration given is:

int ar[3];

This means the array consists of three locations and is named as ar.

We know that the indexes are used to address the locations of an array and the index starts from 0 and goes upto to 1 less than the size of the array which means the indexes of array of 3 elements will start from 0 and end at 2.

Now in the given options we are using ndx variable to run the while loop.

So the code to assign zero to all elements of array will be

ndx = 0;

while(ndx<3)

{

ar[ndx] = 0;

ndx++;

}

Hence, the correct answer is:

C. ndx = 0;

while (ndx < 3) {

ar[ndx] = 0;

ndx++;

}

7 0

3 years ago

Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a

kirill115 [55]

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = $\frac{CPI*Instruction Count}{ClockRate}$

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

for system B CPU time is $=\frac{2.5*100,000}{750*10^6}$

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = = $\frac{2.5*Instruction}{750*10^6}$

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

3 0

3 years ago

Make this a meme please

oksian1 [2.3K]

Dog dog dog dog food

4 0

3 years ago

Read 2 more answers