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In 1985, the average cost of a gallon of milk was [tex]\$2.20[/tex]. In 2023, the average cost is [tex]\$4.10[/tex]. Let [tex]t[/tex] be the number of years since 1980, and [tex]C[/tex] be the cost of a gallon of milk.

Find a linear equation for [tex]C[/tex] in terms of [tex]t[/tex].

A. [tex]C = 0.05t - 97.05[/tex]
B. [tex]C = 20t - 97.8[/tex]
C. [tex]C = 0.05t + 1.95[/tex]
D. [tex]C = 0.05t + 2.20[/tex]
E. [tex]C = 20t + 2.20[/tex]

Answer :

To find a linear equation for the cost of a gallon of milk, [tex]\( C \)[/tex], in terms of the number of years since 1980, [tex]\( t \)[/tex], we'll follow these steps:

1. Identify Key Points:
- In 1985, when [tex]\( t = 5 \)[/tex], the cost [tex]\( C = \$2.20 \)[/tex].
- In 2023, when [tex]\( t = 43 \)[/tex], the cost [tex]\( C = \$4.10 \)[/tex].

2. Calculate the Slope (Rate of Change):
- The slope of a line is given by the change in cost divided by the change in time.
[tex]\[
\text{slope} = \frac{4.10 - 2.20}{43 - 5} = \frac{1.90}{38} = 0.05
\][/tex]

3. Find the Equation of the Line:
- The linear equation has the form [tex]\( C = mt + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- Substitute one of the points to find [tex]\( b \)[/tex]. Let's use [tex]\( t = 5 \)[/tex], [tex]\( C = 2.20 \)[/tex].
[tex]\[
2.20 = 0.05 \times 5 + b
\][/tex]
[tex]\[
2.20 = 0.25 + b
\][/tex]
[tex]\[
b = 2.20 - 0.25 = 1.95
\][/tex]

4. Write the Final Equation:
- Substitute the slope [tex]\( 0.05 \)[/tex] and y-intercept [tex]\( 1.95 \)[/tex] into the equation.
[tex]\[
C = 0.05t + 1.95
\][/tex]

Therefore, the linear equation that describes the cost of a gallon of milk in terms of the number of years since 1980 is [tex]\( C = 0.05t + 1.95 \)[/tex].

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