Rational expectations theory suggests that the speed of adjustment Purcell correction would be very quick.
<h3>What Is Rational Expectations Theory?</h3>
The rational expectations theory is a widely used concept and modeling technique in macroeconomics. Individuals make decisions based on three primary factors, according to the theory: their human rationality, the information available to them, and their past experiences.
The rational expectations hypothesis was originally suggested by John (Jack) Muth 1 (1961) to explain how the outcome of a given economic phenomena depends to a certain degree on what agents expect to happen.
- People who have rational expectations always learn from their mistakes.
- Forecasts are unbiased, and people make decisions based on all available information and economic theories.
- People understand how the economy works and how government policies affect macroeconomic variables like the price level, unemployment rate, and aggregate output.
To learn more about Rational expectations theory from the given link
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Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
Answer:
See explanation
Explanation:
The degradation of the drug is a first order process;
Hence;
ln[A] = ln[A]o - kt
Where;
ln[A] = final concentration of the drug
ln[A]o= initial concentration of the drug = 5 gm/100
k= degradation constant = 0.05 day-1
t= time taken
When [A] =[ A]o - 0.5[A]o = 0.5[A]o
ln2.5 = ln5 - 0.05t
ln2.5- ln5 = - 0.05t
t= ln2.5- ln5/-0.05
t= 0.9162 - 1.6094/-0.05
t= 14 days
b) when [A] = [A]o - 0.9[A]o = 0.1[A]o
ln0.5 = ln5 -0.05t
t= ln0.5 - ln5/0.05
t= -0.693 - 1.6094/-0.05
t= 46 days
Answer:
4
Explanation:
We know that intensity I = P/A where P = power and A = area through which the power passes through.
Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.
So, I = P/A and I₀ = P₀/A
Now, if I = 4I₀,
P/A = 4P₀/A
P = 4P₀
Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t
Substituting P and P₀ into the equation, we have
P = 4P₀
E/t = 4E₀/t
E = 4E₀
Since the energy is four times the initial energy, the energy output increases by a factor of 4.
