Answer :
To determine which of the given situations could be modeled by the equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex], let's analyze each one in detail.
### Given Equation Analysis:
The equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] is a linear equation in slope-intercept form [tex]\( y = mx + b \)[/tex], where:
- [tex]\( m = \frac{5}{2} \)[/tex] is the slope, representing the rate of change.
- [tex]\( b = 20 \)[/tex] is the y-intercept, representing the initial value when [tex]\( x \)[/tex] is zero.
We are looking for situations where there is an initial value of 20 and a rate of change of [tex]\(\frac{5}{2}\)[/tex], which can be represented as [tex]$2.50$[/tex] per unit of [tex]\( x \)[/tex], or some equivalent rate.
### Situation Analysis:
A. Tonya observed a culture of 20 cells. The number of cells in the culture increases by two and a half times each day.
This describes exponential growth, not linear growth. The number of cells increasing by a factor each day is not consistent with the linear equation form.
B. An airport shuttle service charges a \[tex]$20.00 pick-up fee plus an additional \$[/tex]2.50 per mile driven.
- Initial charge (pick-up fee) = \[tex]$20.00 -> This corresponds to \( b = 20 \). - Additional charge per mile = \$[/tex]2.50 -> This corresponds to the slope [tex]\( m = 2.50 \)[/tex].
This situation fits the equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] because the total cost [tex]\( y \)[/tex] is determined by the initial fee plus a rate per mile [tex]\( x \)[/tex].
C. Corey has already walked 20 feet. He will continue to walk at a rate of 5 feet every 2 seconds.
- Initial distance walked = 20 feet -> This matches [tex]\( b = 20 \)[/tex].
- Walking rate = 5 feet every 2 seconds -> This equals [tex]\( \frac{5}{2} \)[/tex] feet per second, which matches the slope [tex]\( m = \frac{5}{2} \)[/tex].
This situation fits the equation because the total distance walked [tex]\( y \)[/tex] increases linearly with the time [tex]\( x \)[/tex].
D. Chef Paulus has baked [tex]\( 21 / 2 \)[/tex] dozen cookies so far. He will continue to bake 20 dozen more per day.
- Initial amount baked = [tex]\( \frac{21}{2} \)[/tex] dozen -> There is some initial value but it does not match 20 exactly.
- Baking rate = 20 dozen per day -> This is not consistent with the slope [tex]\( m = \frac{5}{2} \)[/tex].
This situation does not fit the equation as neither the initial value nor the rate matches our linear model.
E. The temperature is 20 degrees Fahrenheit. The temperature will decrease by [tex]\( 21 / 2 \)[/tex] degrees every hour.
- Initial temperature = 20 degrees -> This matches [tex]\( b = 20 \)[/tex].
- Rate of temperature change = [tex]\(- \frac{21}{2} \)[/tex] degrees per hour (decrease)-> This does not match the slope [tex]\( m = \frac{5}{2} \)[/tex] because the rate is negative and different.
This situation does not fit the equation due to the negative rate of temperature change.
### Conclusion:
The correct situations that can be modeled by the equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] are:
- Situation B: The airport shuttle service.
- Situation C: Corey’s walking scenario.
Therefore, the correct answer is B and C.
### Given Equation Analysis:
The equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] is a linear equation in slope-intercept form [tex]\( y = mx + b \)[/tex], where:
- [tex]\( m = \frac{5}{2} \)[/tex] is the slope, representing the rate of change.
- [tex]\( b = 20 \)[/tex] is the y-intercept, representing the initial value when [tex]\( x \)[/tex] is zero.
We are looking for situations where there is an initial value of 20 and a rate of change of [tex]\(\frac{5}{2}\)[/tex], which can be represented as [tex]$2.50$[/tex] per unit of [tex]\( x \)[/tex], or some equivalent rate.
### Situation Analysis:
A. Tonya observed a culture of 20 cells. The number of cells in the culture increases by two and a half times each day.
This describes exponential growth, not linear growth. The number of cells increasing by a factor each day is not consistent with the linear equation form.
B. An airport shuttle service charges a \[tex]$20.00 pick-up fee plus an additional \$[/tex]2.50 per mile driven.
- Initial charge (pick-up fee) = \[tex]$20.00 -> This corresponds to \( b = 20 \). - Additional charge per mile = \$[/tex]2.50 -> This corresponds to the slope [tex]\( m = 2.50 \)[/tex].
This situation fits the equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] because the total cost [tex]\( y \)[/tex] is determined by the initial fee plus a rate per mile [tex]\( x \)[/tex].
C. Corey has already walked 20 feet. He will continue to walk at a rate of 5 feet every 2 seconds.
- Initial distance walked = 20 feet -> This matches [tex]\( b = 20 \)[/tex].
- Walking rate = 5 feet every 2 seconds -> This equals [tex]\( \frac{5}{2} \)[/tex] feet per second, which matches the slope [tex]\( m = \frac{5}{2} \)[/tex].
This situation fits the equation because the total distance walked [tex]\( y \)[/tex] increases linearly with the time [tex]\( x \)[/tex].
D. Chef Paulus has baked [tex]\( 21 / 2 \)[/tex] dozen cookies so far. He will continue to bake 20 dozen more per day.
- Initial amount baked = [tex]\( \frac{21}{2} \)[/tex] dozen -> There is some initial value but it does not match 20 exactly.
- Baking rate = 20 dozen per day -> This is not consistent with the slope [tex]\( m = \frac{5}{2} \)[/tex].
This situation does not fit the equation as neither the initial value nor the rate matches our linear model.
E. The temperature is 20 degrees Fahrenheit. The temperature will decrease by [tex]\( 21 / 2 \)[/tex] degrees every hour.
- Initial temperature = 20 degrees -> This matches [tex]\( b = 20 \)[/tex].
- Rate of temperature change = [tex]\(- \frac{21}{2} \)[/tex] degrees per hour (decrease)-> This does not match the slope [tex]\( m = \frac{5}{2} \)[/tex] because the rate is negative and different.
This situation does not fit the equation due to the negative rate of temperature change.
### Conclusion:
The correct situations that can be modeled by the equation [tex]\( y = \frac{5}{2} x + 20 \)[/tex] are:
- Situation B: The airport shuttle service.
- Situation C: Corey’s walking scenario.
Therefore, the correct answer is B and C.
