30-[(9x2)-(3x4)]
30-[18-(3x4)]
30-(18-12)
30-6
24
1.5 would be the answer I think
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
A+c=100
3a+2c=275, from the first c=100-a making the 2nd equation become:
3a+2(100-a)=275 perform indicated multiplication on left side
3a+200-2a=275 combine like terms on left side
a+200=275, subtract 200 from both sides
a=75, and since c=100-a
c=100=75=25
So the answer is D. 25 children and 75 adults
Equation 1: a+c=100
Equation 2: 3a+2c=275
Answer:
A. 0.4880
Step-by-step explanation:
First, find the z-score:
z = (x − μ) / σ
z = (13.5 − 13.56) / 2.24
z = -0.03
Using a z-score table:
P(z < -0.03) = 0.4880