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2 years ago

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Computers and Technology

1 answer:

ad-work [718]2 years ago

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Answer:

In the paragraph group, click the Dialog box Launcher. ,and select the Alignment drop - down menu to set your justified text. You can also use the keyboard shortcut , Ctrl + J to justify your text

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Trucking A. is one of the least flexible transportation modes. B. is increasingly using computers to manage its operations. C. i

saw5 [17]

Answer:

B. is increasingly using computers to manage its operations.

Explanation:

Trucking -

It refers to the practice of using computer for the management purpose , is referred to as the process of trucking .

The method is very useful for the business and companies in order to adapt a faster and efficient mode of management .

Hence , from the given information of the question ,

The correct option is b. is increasingly using computers to manage its operations.

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3 years ago

Which of the following statements concerning a short in a series circuit is true?

Romashka [77]

Answer:

what is the answers it gave you?

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2 years ago

The students of a college have to create their assignment reports using a word processing program. Some of the questions in thei

ser-zykov [4K]

Answer:

Table function

Explanation:

The table function can also be used to compare between items. The items can be arranged in columns and the features tonbe compared can be placed in columns. With this one can make comparison between the items.

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2 years ago

Declare k, d, and s so that they can store an integer, a real number, and a small word (under 10 characters). Use these variable

Svetradugi [14.3K]

Answer:

int k;

double d;

char s[10];

cin >> k >> d >> s;

cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s;

Explanation

First Step (declare K, d, s) so they can store a integer

int k;

double d;

char s[10];

Second Step (read in an integer, a real number and a small word)

cin >> k >> d >> s;

Third Step ( print them out )

cout << s << " " << d << " " << k << "\n" << k << " " << d << " " << s;

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3 years ago

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

ANEK [815]

Answer:

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

The exponent bits will be represented as : 100----01.

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

B) The largest odd integer that can be represented exactly

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

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

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

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

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.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

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

The exponent bits will be represented as : 100----01.

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

B) The largest odd integer that can be represented exactly

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

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

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

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

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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3 years ago

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