Which set of points can be modeled without a power model? (2 points)
{(2.2, 4.33), (3.5, 6.641), (4.8, 10.184)}
{(1.4, 0.133), (2, 0.21), (2.6, 0.333)}
{(1.1, 17.155), (1.7, 14.941), (2.3, 13.013)}
{(1.4, 5.192), (1.8,6.104), (2.2, 7.016)}
{(1.4, 5.192), (1.8,6.104), (2.2, 7.016)}
AB
M =( 6.104-5.192)/(1.8-1.4) = 2.28
Bc
M = (7.016-6.104)/(2.2-1.8) = 2.28
because slope is same
To determine whether a set of points can be modeled without a power model, we need to check if the y-values increase or decrease at a consistent rate as the x-values increase. If the y-values increase or decrease exponentially or at varying rates, we may need a power model.
Looking at the given sets of points:
1. {(2.2, 4.33), (3.5, 6.641), (4.8, 10.184)}:
By observing the y-values, we can see that they are not increasing or decreasing at a consistent rate. Therefore, this set of points cannot be modeled without a power model.
2. {(1.4, 0.133), (2, 0.21), (2.6, 0.333)}:
Similar to the previous set, the y-values are also not increasing or decreasing at a consistent rate. Hence, this set of points cannot be modeled without a power model.
3. {(1.1, 17.155), (1.7, 14.941), (2.3, 13.013)}:
By looking at the y-values, it seems that they are decreasing consistently as the x-values increase. Therefore, this set of points can possibly be modeled without a power model.
4. {(1.4, 5.192), (1.8,6.104), (2.2, 7.016)}:
Similarly, the y-values here increase consistently as the x-values increase. So, this set of points can also possibly be modeled without a power model.
In conclusion, the third set of points {(1.1, 17.155), (1.7, 14.941), (2.3, 13.013)} and the fourth set of points {(1.4, 5.192), (1.8,6.104), (2.2, 7.016)} can be modeled without a power model.