A funnel in the shape of an inverted cone is 30 cm. deep and has a diameter across the top of 20 cm. Liquid is flowing out of the funnel at the rate of 12 cm3 /sec. At what rate is the height of the liquid changing at the instant when the liquid in the funnel is 20 cm. deep? Round to the nearest thousandth.

-0.114 cm/sec

Step-by-step explanation:

From the question;

The original radius of the cone is; R = 20/2 = 10 cm

The original height of the cone is; H = 30 cm.

Rate of decrease in the volume of the liquid = 12 cm³/sec, or dV/dt = -12 cm3/sec.

Now, at a later instant, the height of the liquid in the cone is h < H and the radius of the liquid in the cone is r < R. At that time, the volume of liquid is

V = (1/3) πr²h

Now,we know that the ratio of the height and radius of the liquid is the same at any time. So, we can do this by looking at similar triangles;

h/r = H/R

We want to find the rate of height, So, let's make the radius (r) the subject.

Thus;

r = (R/H)h

Substitute this expression for r into the equation for V above. We get

V = (1/3) π(R/H)²h²h = (1/3) π(R/H)h³

Remember R and H are constants, R = 10 cm, H = 30 cm

Now, we can the find the time derivative of each side

Using the chain rule, we have;

dV/dt = (R/H)h²π(dh/dt),

We now want to solve for dh/dt.

Thus,

dh/dt = (H/Rπ)(h^(-2))dV/dt

dV/dt = -12 cm³/sec.

h = 10 cm, since it is 20 cm deep at the time of interest.

So plugging in the relevant values;

dh/dt = (30/10)•(10^(-2))•(1/π) (-12) = -0.114 cm/sec

ghjmutmum

step-by-step explanation:

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5