Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
52 = 50
512= 500
500/50 = 10
How many modes does the data set shown below contain?<br><br><br>
1, 3, 3, 4, 6, 6, 7, 7, 7, 10, 12
kipiarov [429]
Answer:
There are 3 modes
Step-by-step explanation:
There are two 3s, two 6s, and three 7s
Answer:
3^2, 9 students.
Step-by-step explanation:
I'm not 100 percent sure about this but I'm pretty sure that because there are 3 rows you can just do 3^2 which is 9.
Answer:
250 x 3 = 750g
750g is your final answer