Please help me on that question for my Algebra. I don't understand it.
2 answers:
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Answer:
The mean is of -0.4 hours.
Step-by-step explanation:
To solve this question, we need to understand the central limit theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
Mean of the sample of 64 Duracell:
By the Central Limit Theorem, 4.1 hours.
Mean of the sample of 64 Eveready:
By the Central Limit Theorem, 4.5 hours.
Mean of the difference?
Subtraction of normal variables, so we subtract the means.
4.1 - 4.5 = -0.4
The mean is of -0.4 hours.
Answer:
a) (dy/dt) = 2 - [3y/(100 + 2t)]
b) The solved differential equation gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L
Step-by-step explanation:
First of, we take the overall balance for the system,
Let V = volume of solution in the tank at any time
The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)
The rate of change of the volume of solution = dV/dt
Rate of flow into the tank = Fᵢ = 5 L/min
Rate of flow out of the tank = F = 3 L/min
(dV/dt) = Fᵢ - F
(dV/dt) = (Fᵢ - F)
dV = (Fᵢ - F) dt
∫ dV = ∫ (Fᵢ - F) dt
Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t
V - 100 = (Fᵢ - F)t
V = 100 + (5 - 3)t
V = 100 + (2) t
V = (100 + 2t) L
Component balance for the amount of salt in the tank.
Let the initial amount of salt in the tank be y₀ = 0 kg
Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min
Amount of salt in the tank, at any time = y kg
Concentration of salt in the tank at any time = (y/V) kg/L
Recall that V is the volume of water in the tank. V = 100 + 2t
Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min
Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)
(dy/dt) = 2 - (3y/V)
(dy/dt) = 2 - [3y/(100 + 2t)]
To solve this differential equation, it is done in the attached image to this question.
The solution of the differential equation is
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration after 20 minutes.
After 20 minutes, volume of water in tank will be
V(t) = 100 + 2t
V(20) = 100 + 2(20) = 140 L
Amount of salt in the tank after 20 minutes gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵
y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵
y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵
y(20) = 56 - 24.15 = 31.85 kg
Amount of salt in the tank after 20 minutes = 31.85 kg
Volume of water in the tank after 20 minutes = 140 L
Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L
Hope this Helps!!!
The answer is 10i.
look at the picture to solve.
look at the picture to solve.