What sample size would be required if we want to detect a true mean speed as low

1) x  =  [ - (12/1000)*  e∧ ( - 15/1000)*t  + 12/1000 ] /e∧ - (15/1000)*t

2) x =  - 0,012 * ( e ∧ 0.18 + 0,012 ) / e∧-0,18

Step-by-step explanation:

1.-Quantity of salt in the tank after t minutes

The rate  of change of the quantity of salt in the tank is:

dx(t) /dt   =  original quantity (0) + input quantity - output quantity  (1)

quantity = concentration* rate   Then

input quantity   =  0.04 Kg/lt * 5 Lt/min + 0.06 Kg/lt * 10Lt/min  =  0.2 Kg/min

+ 0.6  Kg/min  =  0,8 Kg/lt

output quantity =  Output concentration * rate of draining

rate of draining = 15 Lt/min

The input quantity and the output quantity occur at the same rate therefore the volume in the tank is constant  1000Lt.

output quantity = (x/1000 )*15

Plugging these values in equation (1) we get.

dx/dt  =  0,8  -  ( x/1000)* 15

The last one is a differential first-order equation like

x´ +  P(t)*x = q(t)

and the solution is:

x*μ   =  ∫ q(t)*μ*dt  + C

where    μ  is the integration factor   e ∧ ∫p(t)*dt

let´s call  b = -15/1000

μ = e ∧ ∫p(t)*dt  =  e∧∫ b*dt  =  e∧ b*t = e∧ ( -15/1000)*t

μ  =  e∧ - (15/1000)*t

Then  x*μ    =   x *  e∧ - (15/1000)*t

∫ q(t)*μ*dt  = ∫  0.8 * e∧ - (15/1000)*t*dt  =  0.8 * ∫ e∧bt * dt  

 ∫ q(t)*μ*dt  = 0.8 * ( 1/b ) e∧bt  = - 0,8 *( 15/1000) * e∧ ( - 15/1000)*t

∫ q(t)*μ*dt  = - (12/1000)*  e∧ ( - 15/1000)*t

x *  e∧ - (15/1000)*t  =   - (12/1000)*  e∧ ( - 15/1000)*t + C

Initial condition  t = 0  x = 0

0  =  - (12 / 1000 )* e⁰ = C

C = 12/1000

x *  e∧ - (15/1000)*t  =   - (12/1000)*  e∧ ( - 15/1000)*t  + 12/1000

x  =  [ - (12/1000)*  e∧ ( - 15/1000)*t  + 12/1000 ] /e∧ - (15/1000)*t

When t = 60 min

x =   [ - (12/1000)*  e∧ ( - 15/1000)*12 + 12/1000 ] / e∧ - (15/1000) * 12

x =  - 0,012 * ( e ∧ 0.18 + 0,012 ) / e∧-0,18