The relationship between Celsius $$ \left({ }^{\circ} \mathrm{C}\right) $$ and Fahrenheit $$ \left({ }^{\circ} \mathrm{F}\right) $$ degrees of measuring temperature is linear. Find a linear equation relating $$ { }^{\circ} \mathrm{C} $$ and $$ { }^{\circ} \mathrm{F} $$ if $$ 0^{\circ} \mathrm{C} $$ corresponds to $$ 32^{\circ} \mathrm{F} $$ and $$ 30^{\circ} \mathrm{C} $$ corresponds to $$ 86^{\circ} \mathrm{F} $$. Use the equation to find the Celsius measure of $$ 74^{\circ} \mathrm{F} $$. Choose the correct equation below. A. $$ { }^{\circ} \mathrm{C}=\frac{5}{9}\left({ }^{\circ} \mathrm{F}+32\right) $$ C. $$ { }^{\circ} \mathrm{C}=\frac{9}{5}\left({ }^{\circ} \mathrm{F}-32\right) $$ The temperature of $$ 74^{\circ} \mathrm{F} $$ written as degrees Celsius is $$ \square{ }^{\circ} \mathrm{C} $$. (Round to one decimal place as needed.)

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To establish the linear relationship between Celsius $$ \left(^{\circ} \mathrm{C}\right) $$ and Fahrenheit $$ \left(^{\circ} \mathrm{F}\right) $$, we'll use the two given data points:

1. $$ 0^{\circ} \mathrm{C} = 32^{\circ} \mathrm{F} $$
2. $$ 30^{\circ} \mathrm{C} = 86^{\circ} \mathrm{F} $$

### Step 1: Determine the Linear Equation

Assume the relationship is linear, so it can be expressed as:
$$
^{\circ} \mathrm{C} = m \cdot \, ^{\circ} \mathrm{F} + b
$$
where $$ m $$ is the slope and $$ b $$ is the y-intercept.

Using the first data point:
$$
0 = m \cdot 32 + b \implies b = -32m
$$

Using the second data point:
$$
30 = m \cdot 86 + b \implies 30 = 86m - 32m \implies 30 = 54m \implies m = \frac{5}{9}
$$

Substituting $$ m $$ back into the equation for $$ b $$:
$$
b = -32 \cdot \frac{5}{9} = -\frac{160}{9}
$$

Thus, the linear equation becomes:
$$
^{\circ} \mathrm{C} = \frac{5}{9} \left( \, ^{\circ} \mathrm{F} - 32 \right)
$$

### Step 2: Selecting the Correct Equation

Among the provided options:

- **A.** $$ ^{\circ} \mathrm{C} = \frac{5}{9} \left( \, ^{\circ} \mathrm{F} + 32 \right) $$  
  *This incorrectly adds 32 instead of subtracting.*

- **C.** $$ ^{\circ} \mathrm{C} = \frac{9}{5} \left( \, ^{\circ} \mathrm{F} - 32 \right) $$  
  *This incorrectly uses $$ \frac{9}{5} $$ instead of $$ \frac{5}{9} $$.*

**Neither option A nor C is correct.** The accurate linear equation relating Celsius and Fahrenheit is:
$$
^{\circ} \mathrm{C} = \frac{5}{9} \left( \, ^{\circ} \mathrm{F} - 32 \right)
$$

### Step 3: Convert $$ 74^{\circ} \mathrm{F} $$ to Celsius

Using the correct equation:
$$
^{\circ} \mathrm{C} = \frac{5}{9} \left( 74 - 32 \right) = \frac{5}{9} \times 42 = \frac{210}{9} \approx 23.3^{\circ} \mathrm{C}
$$

**Answer:**  
The correct equation is $$ { }^{\circ} \mathrm{C} = \frac{5}{9}\left({ }^{\circ} \mathrm{F} - 32\right) $$.  
Using this, $$ 74^{\circ} \mathrm{F} $$ is approximately $$ 23.3^{\circ} \mathrm{C} $$.
The correct linear equation relating Celsius and Fahrenheit is: $$ ^{\circ} \mathrm{C} = \frac{5}{9} \left( \, ^{\circ} \mathrm{F} - 32 \right) $$ Using this equation, $$ 74^{\circ} \mathrm{F} $$ is approximately $$ 23.3^{\circ} \mathrm{C} $$.

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