To solve the expression $\frac{4.7620 \times 10^{-15}}{(3.8529 \times 10^{12})(2.813 \times 10^{-7})(9.50)}$ and round the result to the correct significant figures, follow these steps:

Step 1: Calculate the denominator

First, multiply the numbers in the denominator:

$3.8529 \times 10^{12} \times 2.813 \times 10^{-7} \times 9.50$

Performing the multiplications:

$3.8529 \times 2.813 \times 9.50 = 103.05996735$

Now multiply by the powers of ten:

$103.05996735 \times 10^{12} \times 10^{-7} = 103.05996735 \times 10^{5} = 1.0305996735 \times 10^{7}$

Step 2: Divide the numerator by the result from Step 1

Now, divide the numerator by this result:

$\frac{4.7620 \times 10^{-15}}{1.0305996735 \times 10^{7}} = \frac{4.7620}{1.0305996735} \times 10^{-15-7} = \frac{4.7620}{1.0305996735} \times 10^{-22}$

Calculate the division:

$\frac{4.7620}{1.0305996735} \approx 4.620 \times 10^{-22}$

Step 3: Rounding to significant figures

The number $4.7620$ has 5 significant figures, and the denominator multiplication gives us a result where we must round the answer to 4 significant figures, matching the least number of significant figures in the denominator's components (which is 4 from 9.50).

Thus, the final answer is:

$4.620 \times 10^{-22}$

This result is rounded to 4 significant figures, which is consistent with the input numbers.

Would you like more details or have any questions?

Related Questions:

1. How do you determine the number of significant figures in multiplication and division?
2. What is scientific notation, and why is it useful in calculations involving very large or small numbers?
3. How does rounding affect the accuracy of scientific calculations?
4. What is the general rule for rounding numbers in significant figures?
5. How do you handle calculations involving both addition/subtraction and multiplication/division in terms of significant figures?

Tip: Always make sure to match the number of significant figures in your final answer to the smallest number of significant figures in any of the input values.