Consider the following floating-point number.  The mantissa is written in binary, but the 2 and the exponent are in decimal.

0.0000011001100 * 20

If this value is converted to IEEE single precision floating point formate (i.e., a float on the common platform), what binary value will be stored in the exponent field?  Remember that the value will be normalized before it is stored.



To solve this problem, let’s go through the steps required to convert this floating-point number to the IEEE 754 single-precision floating-point format.

### Step 1: Normalize the Floating-Point Number
The given floating-point number is:
\[
0.0000011001100 \times 2^0
\]

In IEEE 754 format, the number needs to be normalized so that it has the form:
\[
1.\text{mantissa} \times 2^{\text{exponent}}
\]

To normalize it, we need to shift the binary point in the mantissa until there's only one non-zero digit (1) to the left of the binary point. 

The original binary mantissa is \( 0.0000011001100 \). Shifting the binary point 6 places to the right gives:
\[
1.1001100 \times 2^{-6}
\]

Thus, the normalized form of the number is:
\[
1.1001100 \times 2^{-6}
\]

### Step 2: Calculate the Exponent in IEEE 754 Format
In IEEE 754 single-precision format, the exponent field is stored using a "bias" of 127. This means the exponent field represents:
\[
\text{Exponent} = \text{actual exponent} + 127
\]

Here, the actual exponent is \(-6\), so we calculate the biased exponent as:
\[
-6 + 127 = 121
\]

### Step 3: Convert the Exponent to Binary
The exponent \(121\) in binary is:
\[
121_{10} = 01111001_2
\]

### Answer
In IEEE 754 single-precision format, the binary value stored in the exponent field for this floating-point number is:
\[
\boxed{01111001}
\]

Would you like further details on any part of the conversion process?

---

1. What is the IEEE 754 format for a double-precision floating point?
2. How do we handle subnormal numbers in IEEE 754?
3. Why is there a bias in the exponent in IEEE floating-point format?
4. How is the mantissa represented in IEEE single-precision format?
5. What are the special values represented by IEEE 754, such as NaN and infinity?

**Tip:** Always remember that IEEE 754 uses a bias in the exponent field to allow both positive and negative exponents.

Consider the following floating-point number. The mantissa is written in binary, but the 2 and the exponent are in decimal.

0.0000011001100 * 2^0

If this value is converted to IEEE single precision floating point format (i.e., a float on the common platform), what binary value will be stored in the exponent field? Remember that the value will be normalized before it is stored.

Learn how to convert the floating-point number 0.0000011001100 * 2^0 into IEEE 754 single-precision format. This guide covers normalization, calculating the biased exponent with a bias of 127, and converting to binary. Ideal for college-level students studying computer science or digital systems.

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