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antiseptic1488 [7]
1 year ago
8

Float_abs - return bit-level equivalent of absolute value of f for * floating point argument f. * both the argument and result a

re passed as unsigned int's, but * they are to be interpreted as the bit-level representations of * single-precision floating point values. * when argument is nan, return argument.. * legal ops: any integer/unsigned operations incl. ||, &&. also if, while * max ops: 10 * rating: 2
Computers and Technology
1 answer:
olga_2 [115]1 year ago
3 0

#include <stdio.h>

int main(void) {

            // your code goes here

            //unsigned a =float_times_four(0x80000000);

            unsigned float_times_four(unsigned uf){

                unsigned expn = (uf >> 23) & 0xFF;

                printf(expn);

                unsigned sign = uf & 0x80000000;

                unsigned frac = uf & 0x007FFFFF;

                if(expn == 255 ||(expn == 0 && frac ==0))

                    return uf;

                if(expn){

                    expn<<2;

                }else if(frac == 0x007FFFFF){

//here 0x7FFFFF given by you that is wrong you place this 0x007FFFFF will excute

                    frac>>2;

                    expn<<2;

                }else{

                    frac<<=2;

                }

return (sign) | (expn <<23) | (frac);

            }

            return 0;

}

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Answer:

The correct answer is:

a. M54.6, C79.51, C80.1

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2 years ago
Given a scanner reference variable named input that has been associated with an input source consisting of a sequence of strings
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7 0
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Read 2 more answers
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Answer:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

Explanation:

Previous concepts

Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"

Solution to the problem

For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:

Available = 4096 MB - 512 MB = 3584 MB

Now we can find the maximum simultaneous process than can use with this:

\frac{3584 MB}{256 MB/proc}= 14 processes

And then we can find the maximum wait I/O that can be tolerated with the following formula:

1- p^{14}= rate

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:

1-p^{14} = 0.99

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3 0
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Answer:

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url_split = url.split('://')

isProtocolValid = False

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url = input("Enter an URL: ")

print(validate_url(url))

Explanation:

The image of the output code is attached. Hope it helps.

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