Delete this answer please.
Answer:
The volume of this rectangular prism is 112mm.
Step-by-step explanation:
Answer:$8.60
I put the work into the picture. Hope this helps!
Answer:
a. Concentration of nitrogen in water draining from fertilized lands
b. Quantitative
c. Water draining from fertilized lands
Step-by-step explanation:
a. Here we are evaluating the Concentration in miligrams of Nitrogen per liter of water, that drains from fertilized lands.. So thats what is defined as the variable.
b. When we talk about qualitative variblaes, we refer to variables that we can't define with numbers. For example the colour of a car, that's a qualitative variable. In this problem, can put a number on the nitrogen concetration. When we can measure the variable with numbers we consider it to be a quantitative variable. Therefore this is a quantitative variable
c. The implied population is the population where we want to interfer the analysis. In here we want to know the concentration of water draining from fertilized lands. So we are using random samples from a lake, and we extrapolate that analysis to a bigger universe, that it´s the water draining from fertilized lands
<h3>Given</h3>
- a cone of height 0.4 m and diameter 0.3 m
- filling at the rate 0.004 m³/s
- fill height of 0.2 m at the time of interest
<h3>Find</h3>
- the rate of change of fill height at the time of interest
<h3>Solution</h3>
The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of
... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²
This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.
... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s
Dividing by the coefficient of dh/dt, we get
... dh/dt = 0.004·16/(0.09π) m/s
... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s
_____
You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)
Note that the cone dimensions mean the radius is 3/8 of the height.
V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³
dV/dt = 9π/64·h²·dh/dt
.004 = 9π/64·0.2²·dh/dt . . . substitute the given values
dh/dt = .004·64/(.04·9·π) = 32/(45π)
