Answer:
must also be positive
Step-by-step explanation:
The sign of the correlation coefficient indicates whether the fitted line is sloping upwards or downwards, so it should be consistent with the sign of the slope of the fitted line. so if the correlation line is poitive so should the regression line
To get this answer you would have to have a pace calculator!
Answer:
0.28cm/min
Step-by-step explanation:
Given the horizontal trough whose ends are isosceles trapezoid
Volume of the Trough =Base Area X Height
=Area of the Trapezoid X Height of the Trough (H)
The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)
The Volume of water in the trough at any time


=8h(8+2x)
V=64h+16hx
We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles
x/h=1/4
4x=h
x=h/4
Substituting x=h/4 into the Volume, V


h=3m,
dV/dt=25cm/min=0.25 m/min

=0.002841m/min =0.28cm/min
The rate is the water being drawn from the trough is 0.28cm/min.