A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) Find the amount of salt in the tank after 1 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)

Step-by-step explanation:

Initial Quantity of salt Q(0) = 100 kg

Capacity of tank = 1000 L

Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

Assume after entering constantly mixes and drains out.

Concentration flow = inflow - outflow

= 0.05(10) -

i.e. , Q(0) = 100

Use Q(0) = 5

-45 = A  

So the equation would be

a) Initial concentratin =

b) Use the solution fo rDE substitute t =1

Q(1) = 50-45e^(-0.01) = 5.4477 kg

c) As t becomes large Q = 50

So concentration limit = 50/1000 = 0.05

answer:

1. 0.04

2. 0.15

step-by-step explanation:

i just finished that assignment for properties of probability distributions.

based on context clues you gave i'm assuming the answer would be c.