Answer :
To determine the time at which the temperature will reach [tex]\( 100^{\circ} C \)[/tex] using the line of best fit, we start by finding the equation of this line in the form [tex]\( y = mx + c \)[/tex], where [tex]\( y \)[/tex] is the temperature, [tex]\( x \)[/tex] is the time in minutes, [tex]\( m \)[/tex] is the slope, and [tex]\( c \)[/tex] is the y-intercept.
The calculations provided earlier give us:
- Slope ([tex]\( m \)[/tex]): 4.539393939393942
- Intercept ([tex]\( c \)[/tex]): 77.83636363636359
The equation of the line of best fit is therefore:
[tex]\[ y = 4.539393939393942x + 77.83636363636359 \][/tex]
We are interested in finding the time [tex]\( x \)[/tex] when the temperature [tex]\( y \)[/tex] reaches [tex]\( 100^{\circ} C \)[/tex]. To do this, we set [tex]\( y \)[/tex] to [tex]\( 100 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 100 = 4.539393939393942x + 77.83636363636359 \][/tex]
To isolate [tex]\( x \)[/tex], follow these steps:
1. Subtract 77.83636363636359 from both sides:
[tex]\[ 100 - 77.83636363636359 = 4.539393939393942x \][/tex]
[tex]\[ 22.16363636363641 = 4.539393939393942x \][/tex]
2. Divide both sides by 4.539393939393942 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{22.16363636363641}{4.539393939393942} \][/tex]
[tex]\[ x \approx 4.882510013351143 \][/tex]
Thus, the temperature is expected to reach [tex]\( 100^{\circ} C \)[/tex] at approximately [tex]\( 4.8825 \)[/tex] minutes.
Given the answer choices:
- 5
- 5.5
- 6
- 6.5
The time [tex]\( 4.8825 \)[/tex] minutes round to the nearest value given in the choices is [tex]\( 5 \)[/tex] minutes.
Therefore, according to the line of best fit, the temperature will reach [tex]\( 100^{\circ} C \)[/tex] at [tex]\( 5 \)[/tex] minutes.
The calculations provided earlier give us:
- Slope ([tex]\( m \)[/tex]): 4.539393939393942
- Intercept ([tex]\( c \)[/tex]): 77.83636363636359
The equation of the line of best fit is therefore:
[tex]\[ y = 4.539393939393942x + 77.83636363636359 \][/tex]
We are interested in finding the time [tex]\( x \)[/tex] when the temperature [tex]\( y \)[/tex] reaches [tex]\( 100^{\circ} C \)[/tex]. To do this, we set [tex]\( y \)[/tex] to [tex]\( 100 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ 100 = 4.539393939393942x + 77.83636363636359 \][/tex]
To isolate [tex]\( x \)[/tex], follow these steps:
1. Subtract 77.83636363636359 from both sides:
[tex]\[ 100 - 77.83636363636359 = 4.539393939393942x \][/tex]
[tex]\[ 22.16363636363641 = 4.539393939393942x \][/tex]
2. Divide both sides by 4.539393939393942 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{22.16363636363641}{4.539393939393942} \][/tex]
[tex]\[ x \approx 4.882510013351143 \][/tex]
Thus, the temperature is expected to reach [tex]\( 100^{\circ} C \)[/tex] at approximately [tex]\( 4.8825 \)[/tex] minutes.
Given the answer choices:
- 5
- 5.5
- 6
- 6.5
The time [tex]\( 4.8825 \)[/tex] minutes round to the nearest value given in the choices is [tex]\( 5 \)[/tex] minutes.
Therefore, according to the line of best fit, the temperature will reach [tex]\( 100^{\circ} C \)[/tex] at [tex]\( 5 \)[/tex] minutes.