Answer :
To predict Denise's length in inches at [tex]\( x \)[/tex] number of months, we will use a linear model of the form:
[tex]\[ L(x) = m \cdot x + b \][/tex]
Where:
- [tex]\( L(x) \)[/tex] is Denise's length at [tex]\( x \)[/tex] months.
- [tex]\( m \)[/tex] is the constant rate of change of Denise's length.
- [tex]\( b \)[/tex] is Denise's length at 0 months.
Given that [tex]\( b \)[/tex] represents Denise's length at 0 months, we need to determine this value to complete the model. Let’s assume we were given two specific data points for Denise's length at two different times to find the constants [tex]\( m \)[/tex] and [tex]\( b \)[/tex].
### Step-by-Step Solution
1. Identify the given data points: Suppose we know Denise's length at two different months, say:
- At month [tex]\( x_1 = 2 \)[/tex], Denise's length [tex]\( L(x_1) = 25 \)[/tex] inches.
- At month [tex]\( x_2 = 5 \)[/tex], Denise's length [tex]\( L(x_2) = 34 \)[/tex] inches.
2. Calculate the rate of change, [tex]\( m \)[/tex]:
The rate of change [tex]\( m \)[/tex] is the slope of the line connecting the two points. We can calculate it using the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ m = \frac{L(x_2) - L(x_1)}{x_2 - x_1} \][/tex]
Plugging in our values:
[tex]\[ m = \frac{34 - 25}{5 - 2} = \frac{9}{3} = 3 \][/tex]
So, the rate of change, [tex]\( m \)[/tex], is 3 inches per month.
3. Use one of the data points to solve for [tex]\( b \)[/tex]:
We know that at [tex]\( x = 2 \)[/tex], [tex]\( L(x) = 25 \)[/tex]. Plugging these values into our linear model [tex]\( L(x) = mx + b \)[/tex]:
[tex]\[ 25 = 3 \cdot 2 + b \][/tex]
Simplifying:
[tex]\[ 25 = 6 + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = 25 - 6 = 19 \][/tex]
So, Denise's length at 0 months, [tex]\( b \)[/tex], is 19 inches.
### Predicting Denise's Length
With the values of [tex]\( m \)[/tex] and [tex]\( b \)[/tex] determined, the function equation to predict Denise's length at [tex]\( x \)[/tex] months is:
[tex]\[ L(x) = 3x + 19 \][/tex]
Thus, using this model, we can predict Denise's length at any number of months by substituting [tex]\( x \)[/tex] with the required month value.
[tex]\[ L(x) = m \cdot x + b \][/tex]
Where:
- [tex]\( L(x) \)[/tex] is Denise's length at [tex]\( x \)[/tex] months.
- [tex]\( m \)[/tex] is the constant rate of change of Denise's length.
- [tex]\( b \)[/tex] is Denise's length at 0 months.
Given that [tex]\( b \)[/tex] represents Denise's length at 0 months, we need to determine this value to complete the model. Let’s assume we were given two specific data points for Denise's length at two different times to find the constants [tex]\( m \)[/tex] and [tex]\( b \)[/tex].
### Step-by-Step Solution
1. Identify the given data points: Suppose we know Denise's length at two different months, say:
- At month [tex]\( x_1 = 2 \)[/tex], Denise's length [tex]\( L(x_1) = 25 \)[/tex] inches.
- At month [tex]\( x_2 = 5 \)[/tex], Denise's length [tex]\( L(x_2) = 34 \)[/tex] inches.
2. Calculate the rate of change, [tex]\( m \)[/tex]:
The rate of change [tex]\( m \)[/tex] is the slope of the line connecting the two points. We can calculate it using the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ m = \frac{L(x_2) - L(x_1)}{x_2 - x_1} \][/tex]
Plugging in our values:
[tex]\[ m = \frac{34 - 25}{5 - 2} = \frac{9}{3} = 3 \][/tex]
So, the rate of change, [tex]\( m \)[/tex], is 3 inches per month.
3. Use one of the data points to solve for [tex]\( b \)[/tex]:
We know that at [tex]\( x = 2 \)[/tex], [tex]\( L(x) = 25 \)[/tex]. Plugging these values into our linear model [tex]\( L(x) = mx + b \)[/tex]:
[tex]\[ 25 = 3 \cdot 2 + b \][/tex]
Simplifying:
[tex]\[ 25 = 6 + b \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = 25 - 6 = 19 \][/tex]
So, Denise's length at 0 months, [tex]\( b \)[/tex], is 19 inches.
### Predicting Denise's Length
With the values of [tex]\( m \)[/tex] and [tex]\( b \)[/tex] determined, the function equation to predict Denise's length at [tex]\( x \)[/tex] months is:
[tex]\[ L(x) = 3x + 19 \][/tex]
Thus, using this model, we can predict Denise's length at any number of months by substituting [tex]\( x \)[/tex] with the required month value.