Answer :
To find the linear relationship between the time of day [tex]\( t \)[/tex] and the temperature [tex]\( F \)[/tex] in degrees Fahrenheit, we can use the form of a linear equation [tex]\( F = mt + b \)[/tex].
In this problem, we are given two data points:
- At 6:00 A.M. ( [tex]\( t = 6 \)[/tex]), the temperature is [tex]\( F = 59 \)[/tex]
- At 10:00 A.M. ( [tex]\( t = 10 \)[/tex]), the temperature is [tex]\( F = 73 \)[/tex]
### Step 1: Determine the Slope ( [tex]\( m \)[/tex] )
The slope [tex]\( m \)[/tex] of a line through two points [tex]\( (t_1, F_1) \)[/tex] and [tex]\( (t_2, F_2) \)[/tex] is given by the formula:
[tex]\[ m = \frac{F_2 - F_1}{t_2 - t_1} \][/tex]
Using the given data points:
[tex]\[ (t_1, F_1) = (6, 59) \][/tex]
[tex]\[ (t_2, F_2) = (10, 73) \][/tex]
Substitute these values into the formula:
[tex]\[ m = \frac{73 - 59}{10 - 6} \][/tex]
[tex]\[ m = \frac{14}{4} \][/tex]
[tex]\[ m = 3.5 \][/tex]
### Step 2: Determine the Y-Intercept ( [tex]\( b \)[/tex] )
The linear equation can be expressed as:
[tex]\[ F = mt + b \][/tex]
To find [tex]\( b \)[/tex], we can use one of our data points. Using [tex]\( (6, 59) \)[/tex]:
[tex]\[ 59 = 3.5 \cdot 6 + b \][/tex]
Solve for [tex]\( b \)[/tex]:
[tex]\[ 59 = 21 + b \][/tex]
[tex]\[ b = 59 - 21 \][/tex]
[tex]\[ b = 38 \][/tex]
### Step 3: Write the Equation
Now that we have determined the slope [tex]\( m = 3.5 \)[/tex] and the y-intercept [tex]\( b = 38 \)[/tex], we can write the linear relationship between [tex]\( t \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ F = 3.5t + 38 \][/tex]
Thus, the linear relationship between the time of the day [tex]\( t \)[/tex] and the temperature [tex]\( F \)[/tex] is:
[tex]\[ F = 3.5t + 38 \][/tex]
This equation represents how the temperature in degrees Fahrenheit depends linearly on the time of the day [tex]\( t \)[/tex].
In this problem, we are given two data points:
- At 6:00 A.M. ( [tex]\( t = 6 \)[/tex]), the temperature is [tex]\( F = 59 \)[/tex]
- At 10:00 A.M. ( [tex]\( t = 10 \)[/tex]), the temperature is [tex]\( F = 73 \)[/tex]
### Step 1: Determine the Slope ( [tex]\( m \)[/tex] )
The slope [tex]\( m \)[/tex] of a line through two points [tex]\( (t_1, F_1) \)[/tex] and [tex]\( (t_2, F_2) \)[/tex] is given by the formula:
[tex]\[ m = \frac{F_2 - F_1}{t_2 - t_1} \][/tex]
Using the given data points:
[tex]\[ (t_1, F_1) = (6, 59) \][/tex]
[tex]\[ (t_2, F_2) = (10, 73) \][/tex]
Substitute these values into the formula:
[tex]\[ m = \frac{73 - 59}{10 - 6} \][/tex]
[tex]\[ m = \frac{14}{4} \][/tex]
[tex]\[ m = 3.5 \][/tex]
### Step 2: Determine the Y-Intercept ( [tex]\( b \)[/tex] )
The linear equation can be expressed as:
[tex]\[ F = mt + b \][/tex]
To find [tex]\( b \)[/tex], we can use one of our data points. Using [tex]\( (6, 59) \)[/tex]:
[tex]\[ 59 = 3.5 \cdot 6 + b \][/tex]
Solve for [tex]\( b \)[/tex]:
[tex]\[ 59 = 21 + b \][/tex]
[tex]\[ b = 59 - 21 \][/tex]
[tex]\[ b = 38 \][/tex]
### Step 3: Write the Equation
Now that we have determined the slope [tex]\( m = 3.5 \)[/tex] and the y-intercept [tex]\( b = 38 \)[/tex], we can write the linear relationship between [tex]\( t \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ F = 3.5t + 38 \][/tex]
Thus, the linear relationship between the time of the day [tex]\( t \)[/tex] and the temperature [tex]\( F \)[/tex] is:
[tex]\[ F = 3.5t + 38 \][/tex]
This equation represents how the temperature in degrees Fahrenheit depends linearly on the time of the day [tex]\( t \)[/tex].
