Answer :
Let's analyze the problem and perform a step-by-step solution.
1. Identify the Line of Best Fit Equation:
The line of best fit is given by the equation:
[tex]\[ h = -21.962x + 114.655 \][/tex]
where [tex]\( h \)[/tex] represents the height of the object and [tex]\( x \)[/tex] represents the time in seconds.
2. Calculate the Predicted Height at 3.5 seconds:
Substitute [tex]\( x = 3.5 \)[/tex] into the line of best fit equation to find the predicted height:
[tex]\[ h = -21.962(3.5) + 114.655 \][/tex]
This results in a predicted height of approximately 37.788 meters.
3. Compare Predicted Height with Actual Height at 3.5 seconds:
From the table, the actual height at 3.5 seconds is 40 meters.
Therefore, the predicted height (37.788 meters) is slightly different from the actual height (40 meters).
4. Determine When the Line of Best Fit Hits the Ground:
The object hits the ground when its height [tex]\( h = 0 \)[/tex]. Set up the equation:
[tex]\[ 0 = -21.962x + 114.655 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 21.962x = 114.655 \][/tex]
[tex]\[ x \approx \frac{114.655}{21.962} \approx 5.2206 \text{ seconds} \][/tex]
According to the line of best fit, the object hits the ground at approximately 5.2206 seconds.
5. Compare Actual Time with Predicted Time of Hitting the Ground:
From the table, the actual time the object hits the ground is 4.6 seconds.
The time difference is:
[tex]\[ 4.6 - 5.2206 \approx -0.6206 \text{ seconds} \][/tex]
This indicates the object actually hit the ground approximately 0.6206 seconds earlier than the line of best fit predicts.
6. Check the Given Statements:
- Statement 1: "According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground."
This is not correct because the correct time difference is approximately -0.6206 seconds, not 0.6 seconds.
- Statement 2: "According to the line of best fit, the object was dropped from a lower height."
This is not mentioned explicitly in the problem. The initial height from the line of best fit at [tex]\( x=0 \)[/tex] is 114.655 meters, which is higher than the actual initial height of 100 meters.
- Statement 3: "The line of best fit correctly predicts that the object reaches a height of 40 meters after 3.5 seconds."
The predicted height at 3.5 seconds is approximately 37.788 meters, not exactly 40 meters. Hence, this statement is not correct.
- Statement 4: "The line of best fit predicts a height of 4 meters greater than the actual height for any time given in the table."
This is not true as the line of best fit doesn't predict a consistent 4 meters' difference over all times in the table.
Thus, none of the provided statements are entirely correct based on the calculations given.
1. Identify the Line of Best Fit Equation:
The line of best fit is given by the equation:
[tex]\[ h = -21.962x + 114.655 \][/tex]
where [tex]\( h \)[/tex] represents the height of the object and [tex]\( x \)[/tex] represents the time in seconds.
2. Calculate the Predicted Height at 3.5 seconds:
Substitute [tex]\( x = 3.5 \)[/tex] into the line of best fit equation to find the predicted height:
[tex]\[ h = -21.962(3.5) + 114.655 \][/tex]
This results in a predicted height of approximately 37.788 meters.
3. Compare Predicted Height with Actual Height at 3.5 seconds:
From the table, the actual height at 3.5 seconds is 40 meters.
Therefore, the predicted height (37.788 meters) is slightly different from the actual height (40 meters).
4. Determine When the Line of Best Fit Hits the Ground:
The object hits the ground when its height [tex]\( h = 0 \)[/tex]. Set up the equation:
[tex]\[ 0 = -21.962x + 114.655 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 21.962x = 114.655 \][/tex]
[tex]\[ x \approx \frac{114.655}{21.962} \approx 5.2206 \text{ seconds} \][/tex]
According to the line of best fit, the object hits the ground at approximately 5.2206 seconds.
5. Compare Actual Time with Predicted Time of Hitting the Ground:
From the table, the actual time the object hits the ground is 4.6 seconds.
The time difference is:
[tex]\[ 4.6 - 5.2206 \approx -0.6206 \text{ seconds} \][/tex]
This indicates the object actually hit the ground approximately 0.6206 seconds earlier than the line of best fit predicts.
6. Check the Given Statements:
- Statement 1: "According to the line of best fit, the object would have hit the ground 0.6 seconds later than the actual time the object hit the ground."
This is not correct because the correct time difference is approximately -0.6206 seconds, not 0.6 seconds.
- Statement 2: "According to the line of best fit, the object was dropped from a lower height."
This is not mentioned explicitly in the problem. The initial height from the line of best fit at [tex]\( x=0 \)[/tex] is 114.655 meters, which is higher than the actual initial height of 100 meters.
- Statement 3: "The line of best fit correctly predicts that the object reaches a height of 40 meters after 3.5 seconds."
The predicted height at 3.5 seconds is approximately 37.788 meters, not exactly 40 meters. Hence, this statement is not correct.
- Statement 4: "The line of best fit predicts a height of 4 meters greater than the actual height for any time given in the table."
This is not true as the line of best fit doesn't predict a consistent 4 meters' difference over all times in the table.
Thus, none of the provided statements are entirely correct based on the calculations given.
