A car company wants to predict how much fuel different cars will use based on their masses. They took a sample of cars, drove ea
ch car 100\,\text{km}100km100, start text, k, m, end text, and measured how much fuel was used in each case (in liters). The data is summarized in the table below.
Fuel used (L) 3.6, 6.7, 9.8, 11.2, 14.7
Mass (metric tons) 0.45, 0.91, 1.36, 1.81, 2.27
All of the scatter plots below display the data correctly, but which one of them displays the data best?
By convention, a good scatter plot uses a reasonable scale on both axes and puts the explanatory variable on the x-axis.
Choose 1 answer:
(Choice A)
(Choice B)
(Choice C)
(Choice D)
Fuel used (L) 3.6, 6.7, 9.8, 11.2, 14.7
Mass (metric tons) 0.45, 0.91, 1.36, 1.81, 2.27
All of the scatter plots below display the data correctly, but which one of them displays the data best?
By convention, a good scatter plot uses a reasonable scale on both axes and puts the explanatory variable on the x-axis.
Choose 1 answer:
(Choice A)
(Choice B)
(Choice C)
(Choice D)
2 answers:
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Answer:
a. The rock's velocity is and the acceleration is
b. It takes 22.5 seconds to reach the highest point.
c. The rock goes up to 405 m.
d. It reach half its maximum height when time is 6.59 s or 38.41 s.
e. The rock is aloft for 45 seconds.
Step-by-step explanation:
- Velocity is defined as the rate of change of position or the rate of displacement.
- Acceleration is defined as the rate of change of velocity.
a.
The rock's velocity is the derivative of the height function
The rock's acceleration is the derivative of the velocity function
b. The rock will reach its highest point when the velocity becomes zero.
It takes 22.5 seconds to reach the highest point.
c. The rock reach its highest point when t = 22.5 s
Thus
So the rock goes up to 405 m.
d. The maximum height is 405 m. So the half of its maximum height =
To find the time it reach half its maximum height, we need to solve
For a quadratic equation of the form the solutions are
It reach half its maximum height when time is 6.59 s or 38.41 s.
e. It is aloft until s(t) = 0 again
The rock is aloft for 45 seconds.