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?

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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.