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To determine the time at which the temperature of the water will reach [tex]\(100^\circ C\)[/tex] based on the given data points, we need to find the equation of the line of best fit that represents the relationship between time and temperature.
1. Finding the line of best fit: The line of best fit is typically represented by the equation of a straight line in the form [tex]\( y = mx + c \)[/tex], where:
- [tex]\( y \)[/tex] represents the temperature.
- [tex]\( x \)[/tex] represents the time.
- [tex]\( m \)[/tex] is the slope of the line.
- [tex]\( c \)[/tex] is the y-intercept.
2. Slope (m) and Intercept (c):
- The slope ([tex]\(m\)[/tex]) of the line is the rate of change of temperature with respect to time.
- The y-intercept ([tex]\(c\)[/tex]) is the temperature at time [tex]\( x = 0 \)[/tex].
Given the numerical results:
- Slope ([tex]\(m\)[/tex]): 4.539393939393942
- Intercept ([tex]\(c\)[/tex]): 77.83636363636359
3. Using the equation of the line of best fit:
[tex]\[ \text{Temperature} = \text{Slope} \times \text{Time} + \text{Intercept} \][/tex]
Plugging in the values:
[tex]\[ T = 4.539393939393942 \times t + 77.83636363636359 \][/tex]
4. Finding the time when the temperature reaches [tex]\(100^\circ C\)[/tex]:
- We set [tex]\( T = 100 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 100 = 4.539393939393942 \times t + 77.83636363636359 \][/tex]
5. Isolating [tex]\( t \)[/tex]:
[tex]\[ 100 - 77.83636363636359 = 4.539393939393942 \times t \][/tex]
[tex]\[ 22.16363636363641 = 4.539393939393942 \times t \][/tex]
[tex]\[ t = \frac{22.16363636363641}{4.539393939393942} \][/tex]
[tex]\[ t \approx 4.882510013351143 \][/tex]
Therefore, the time when the temperature reaches 100°C is approximately [tex]\(4.88\)[/tex] minutes.
Conclusion: Since the provided choices are in whole and half minutes:
- The closest value to [tex]\( 4.88 \)[/tex] minutes is [tex]\( 5 \)[/tex] minutes.
Thus, the temperature will reach [tex]\( 100^\circ C \)[/tex] at approximately [tex]\( 5 \)[/tex] minutes.
1. Finding the line of best fit: The line of best fit is typically represented by the equation of a straight line in the form [tex]\( y = mx + c \)[/tex], where:
- [tex]\( y \)[/tex] represents the temperature.
- [tex]\( x \)[/tex] represents the time.
- [tex]\( m \)[/tex] is the slope of the line.
- [tex]\( c \)[/tex] is the y-intercept.
2. Slope (m) and Intercept (c):
- The slope ([tex]\(m\)[/tex]) of the line is the rate of change of temperature with respect to time.
- The y-intercept ([tex]\(c\)[/tex]) is the temperature at time [tex]\( x = 0 \)[/tex].
Given the numerical results:
- Slope ([tex]\(m\)[/tex]): 4.539393939393942
- Intercept ([tex]\(c\)[/tex]): 77.83636363636359
3. Using the equation of the line of best fit:
[tex]\[ \text{Temperature} = \text{Slope} \times \text{Time} + \text{Intercept} \][/tex]
Plugging in the values:
[tex]\[ T = 4.539393939393942 \times t + 77.83636363636359 \][/tex]
4. Finding the time when the temperature reaches [tex]\(100^\circ C\)[/tex]:
- We set [tex]\( T = 100 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 100 = 4.539393939393942 \times t + 77.83636363636359 \][/tex]
5. Isolating [tex]\( t \)[/tex]:
[tex]\[ 100 - 77.83636363636359 = 4.539393939393942 \times t \][/tex]
[tex]\[ 22.16363636363641 = 4.539393939393942 \times t \][/tex]
[tex]\[ t = \frac{22.16363636363641}{4.539393939393942} \][/tex]
[tex]\[ t \approx 4.882510013351143 \][/tex]
Therefore, the time when the temperature reaches 100°C is approximately [tex]\(4.88\)[/tex] minutes.
Conclusion: Since the provided choices are in whole and half minutes:
- The closest value to [tex]\( 4.88 \)[/tex] minutes is [tex]\( 5 \)[/tex] minutes.
Thus, the temperature will reach [tex]\( 100^\circ C \)[/tex] at approximately [tex]\( 5 \)[/tex] minutes.
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