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vovikov84 [41]
3 years ago
5

Consider the following code segment. int[] seq = {3, 1, 8, 4, 2, 5}; for (int k = 1; k < seq.length; k++) { if (seq[k] >=

seq[0]) System.out.print(seq[k] + " " + k + " "); } What will be printed as a result of executing the code segment?
Computers and Technology
1 answer:
Setler79 [48]3 years ago
4 0

Answer:

This program will complete in 5 iterations,

initially

Seq[0]=3

Iteration 1:

loop starting from 1 and ends at 5 so

when  K=1,   then seq[1]=1

if seq[1]>=Seq[0]    ==>     1 is not greater than equal to 3

so "nothing to print because condition not true"

Iteration 2:

Now

K=2 so seq[2]=8

if seq[2]>=Seq[0]    ==>     8 is greater than equal to 3

so "condition is true" the output will be

output="8","2" means seq[k]=8 and K=2

Iteration 3:

Now

K=3 so seq[3]=4

if seq[3]>=Seq[0]    ==>     4 is greater than equal to 3

so "condition is true" the output will be

output="4","3"  means seq[k]=4 and K=3

Iteration 4:

Now

K=4 so seq[4]=2

if seq[4]>=Seq[0]    ==>     2 is not greater than equal to 3

so "condition is not true" the output will be

output=nothing

Iteration 5:

Now

K=5 so seq[5]=5

if seq[5]>=Seq[0]    ==>     5 is greater than equal to 3

so "condition is true" the output will be

output="5","5"    means seq[k]=5 and K=5

Explanation:

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residuals are A. data collected from individuals that is not consistent with the rest of the group. B. the difference between ob
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Answer:

Option B is correct.

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Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th
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E = integer value of exponent

M = significand value

F = fractional value

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A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112  ≈ 7/4

fractional value ( F )  = 0.112

And, numeric value of the quantity ( V )  = 7

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while The fraction bits will be represented  as : 1100---0.

<u>B) The largest odd integer that can be represented exactly </u>

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

=  1.11------12 = 2 - 2-n

also the fractional value ( F ) =  

0.11------12 = 1 – 2-n

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while The bit representation for the fraction will be as follows: 11---11.

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The exponential value ( E )  = 2k-1 – 2

While the significand value ( M )  = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

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