Question

question 8 (6 points)
which is the correct conversion factor for converting:

7,289dg x ____ = ____ g

○ a $\frac{1g}{1000dg}$
○ b $\frac{10g}{1dg}$
○ c $\frac{1g}{10dg}$
○ d $\frac{10dg}{1g}$

Explanation:

Step1: Recall unit conversion (dg to g)

We know that \( 1\,\text{g} = 10\,\text{dg} \), so the conversion factor to convert decigrams (dg) to grams (g) should have grams in the numerator and decigrams in the denominator, or vice - versa such that the units cancel correctly. To convert \( \text{dg} \) to \( \text{g} \), we use the factor \( \frac{1\,\text{g}}{10\,\text{dg}} \) because when we multiply by this factor, the \( \text{dg} \) units will cancel out, leaving us with grams.

Let's analyze each option:

• Option a: \( \frac{1\,\text{g}}{1000\,\text{dg}} \) is incorrect because \( 1\,\text{g}

eq1000\,\text{dg} \).

• Option b: The expression \( \frac{10\,\text{g}}{10\,\text{g}} \) simplifies to 1, but it does not convert \( \text{dg} \) to \( \text{g} \) as the units of \( \text{g} \) are not related to \( \text{dg} \) in this way. Also, the notation seems incorrect.
• Option c: \( \frac{1\,\text{g}}{10\,\text{dg}} \) is correct. If we multiply \( 7289\,\text{dg} \) by \( \frac{1\,\text{g}}{10\,\text{dg}} \), we get \( 7289\,\text{dg}\times\frac{1\,\text{g}}{10\,\text{dg}}=\frac{7289}{10}\,\text{g} = 728.9\,\text{g} \), and the \( \text{dg} \) units cancel.
• Option d: \( \frac{10\,\text{dg}}{1\,\text{g}} \) would convert grams to decigrams, not the other way around. If we multiply \( 7289\,\text{dg} \) by \( \frac{10\,\text{dg}}{1\,\text{g}} \), we would get units of \( \text{dg}^2/\text{g} \), which is not what we want.

Step2: Calculate the result (optional for the conversion factor question, but to verify)

Using the correct conversion factor from option c: \( 7289\,\text{dg}\times\frac{1\,\text{g}}{10\,\text{dg}}=\frac{7289}{10}\,\text{g}=728.9\,\text{g} \)

Answer:

c. \(\frac{1\,\text{g}}{10\,\text{dg}}\)