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Ugo [173]
3 years ago
8

Does anyone have code hs circle pyramid 2.0?

Computers and Technology
1 answer:
Andrews [41]3 years ago
8 0

Answer:i dont

Explanation:

You might be interested in
Please help me!! 10 Points are waiting!!
prisoha [69]

Google

https://google.com

Google has a large index of keywords and where those words can be found. Google uses a trademarked algorithm called PageRank, which assigns each Web page a relevancy score.

3 0
3 years ago
Design and implement an application that reads a string from the user then determines and prints how many of eachlowercase vowel
frez [133]
Design and implement an application that reads a string from the user, then determines and prints how many of each lowercase vowel (a, e, i, o, and u) appear in the entire string . Have a separate counter for each vowel. Also count and print the number of nonvowel characters .

SPECIFICATION OF PROMPTS, LABELS AND OUTPUT : Your code should use the prompt "enter string : ". After the input is read, there are six lines of output , each starting with a different label: "a: ", "e: ", "i: ", "o: ", "u: ", "other: " in that order. After each label is the required count.

For example: if "aardvark heebie jeebies" were read in,
4 0
3 years ago
What is meant by saving a sheet document​
Zina [86]

Answer:

While you'll use Save or press Ctrl+S to save an existing workbook in its current location, you need to use Save As to save your workbook for the first time, in a different location, or to create a copy of your workbook in the same or another location.

5 0
3 years ago
Match the description to the step in troubleshooting.
Digiron [165]
It should go 4 5 3 2 1
5 0
4 years ago
. Suppose an instruction takes 1/2 microsecond to execute (on the average), and a page fault takes 250 microseconds of processor
saul85 [17]

Answer:

a. 100.

b. 31500.

Explanation:

So, we are given the following data which is going to help in solving this particular question.

The time required to execute (on the average) = 1/2 microsecond , a page fault takes of processor time to handle = 250 microseconds and the disk time to read in the page = 10 milliseconds.

Thus, the time taken by the processor to handle the page fault = 250 microseconds / 1000 = 0.25 milliseconds.

The execution time = [ 1/2 microseconds ]/ 1000 = 0.0005 milliseconds.

The number of Pages sent in a second by the disc = 1000/10 milliseconds = 100.

Assuming U = 1.

Hence, the disc transfer time = [2/3 × 1 } + [ 1/3 × 0.25 milliseconds + 15 ] × 2.

=0.667 + 15.083.

= 15.75 millisecond.

Average number of instruction = 15.75/0.0005 = 31500.

8 0
3 years ago
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