Certainly! Let's tackle each part of the question with detailed, step-by-step explanations.
### Scenario a: Exponential Growth
1. Initial Information:
- Initial cost of a gallon of milk: [tex]$2.00
- Monthly increase rate: 6.1% (which is 0.061 in decimal form)
2. Exponential Equation to Model the Cost Over Time:
- The general form of an exponential growth equation is \(C = C_0 (1 + r)^x\), where:
- \(C\) is the cost after \(x\) months.
- \(C_0\) is the initial cost.
- \(r\) is the monthly rate of increase.
- \(x\) is the number of months.
- Substituting the given values, we have:
\[
C = 2.00 (1 + 0.061)^x
\]
### Scenario b: Linear Growth
1. Initial Information:
- Initial cost of a gallon of milk: $[/tex]2.00
- Monthly increase amount: [tex]$0.138
2. Linear Equation to Model the Cost Over Time:
- The general form of a linear growth equation is \(C = C_0 + mx\), where:
- \(C\) is the cost after \(x\) months.
- \(C_0\) is the initial cost.
- \(m\) is the monthly increase amount.
- \(x\) is the number of months.
- Substituting the given values, we have:
\[
C = 2.00 + 0.138x
\]
### Predicting the Cost After 12 Months:
Next, we'll use both equations to predict the cost of a gallon of milk 12 months from July, which means \(x = 12\).
#### For the Exponential Growth Equation:
\[
C_{\text{exponential}} = 2.00 (1 + 0.061)^{12}
\]
After evaluating the expression, we know the cost after 12 months will be:
\[
C_{\text{exponential}} = 4.070189245918128 \approx 4.07
\]
So, the predicted cost of a gallon of milk in July (a year later) using the exponential model is approximately \$[/tex]4.07.
#### For the Linear Growth Equation:
[tex]\[
C_{\text{linear}} = 2.00 + 0.138 \times 12
\][/tex]
After evaluating the expression, we know the cost after 12 months will be:
[tex]\[
C_{\text{linear}} = 2.00 + 1.656 = 3.656 \approx 3.66
\][/tex]
So, the predicted cost of a gallon of milk in July (a year later) using the linear model is approximately \[tex]$3.66.
### Summary
- Linear Prediction: \$[/tex]3.66
- Exponential Prediction: \$4.07
