Fractional binary number for given hexadecimal representation: From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”. Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”. Hence, the floating point representation for given binary value is From the above table, it is based on the normalized case. Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is Formula for computing “E”, “E = e – bias” Formula for computing “M”, “M = 1 + f” Formula for computing “V”, “ V = ( − 1) s × 2 E × M ” From the above table, sign bit “s” is “0”. E = e − bias = 128 − bias = > Computing decimal value for exponent bit "10000000" is 128 = 128 − ( 2 8 − 1 − 1 ) = > Formula determine bias is 2 k − 1 − 1

Question

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Answer 1

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Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

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Answer 2

Expert Solution & Answer

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

Answer 8

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

Answer 9

Chapter 2, Problem 2.91HW

A.

Explanation of Solution

Fractional binary number for given hexadecimal representation:

From the given question, the binary value for given hexadecimal is “0x40490FDB” is “0100 0000 0100 1001 0000 1111 1101 1011”.

Using the single precision floating point approximation, the bit of “sign” is “1”, for exponent bit “k” is “8” and for fraction bits is “23”.

Hence, the floating point representation for given binary value is

From the above table, it is based on the normalized case.

Because, the value of exponent is not all zeros. Then formula for computing “E”, “M” and “V” is

Formula for computing “E”, “E = e – bias”

Formula for computing “M”, “M = 1 + f”

Formula for computing “V”, “V = (−1)s × 2E × M”

From the above table, sign bit “s” is “0”.

E = e − bias = 128 − bias => Computing decimal value for exponent bit "10000000" is 128 = 128 − (28−1 − 1) => Formula determine bias is 2k−1 − 1

Answer 10

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

Answer 11

B.

Explanation of Solution

Fractional binary representation of “227”:

From problem 2.83, assume n = 0.yyyyyy. So, from part A, n = 11.00 1 001 000 011 111 101 101 1 . Hence, the value of “y” is “001”.

Now formula to computing numeric value for string “001” is given below

n = Y2k − 1

Here the value of “Y” means decimal value for “001” and “k” means number of bits in given binary “001”

Answer 12

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

Answer 13

C.

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

Answer 14

Expert Solution & Answer

Answer 15

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Answer 16

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Answer 17

Ch. 2.1 - Practice Problem 2.1 (solution page 143) Perform...Ch. 2.1 - Prob. 2.2PP Ch. 2.1 - Prob. 2.3PP Ch. 2.1 - Practice Problem 2.4 (solution page 144) Without...Ch. 2.1 - Prob. 2.5PP Ch. 2.1 - Prob. 2.6PP Ch. 2.1 - Prob. 2.7PP Ch. 2.1 - Prob. 2.8PP Ch. 2.1 - Practice Problem 2.9 solution page 146 Computers...Ch. 2.1 - Prob. 2.10PP

Solution Summary: The author explains that the floating point representation for given binary value is based on the normalized case.

Explanation of Solution

Explanation of Solution

Explanation of Solution

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3

Want to see the full answer?

Chapter 2 Solutions

Solution Summary: The author explains that the floating point representation for given binary value is based on the normalized case.

Explanation of Solution

Explanation of Solution

Explanation of Solution

Bit positions of “π” diverge from two approximations: Relative to the binary point, at the 9th bit position, the “π” diverges from two approximations of “22371” and “227”. When converting “22371” to decimal, the value is “3

Want to see the full answer?

Chapter 2 Solutions

Bit positions of “π” diverge from two approximations:

Relative to the binary point, at the 9^th bit position, the “π” diverges from two approximations of “22371” and “227”.

When converting “22371” to decimal, the value is “3