Answer:
N = 920(1+0.03)^4t
Step-by-step explanation:
According to the given statement a car repair center services 920 cars in 2012. The number of cars serviced increases quarterly at a rate of 12% per year after 2012.
Rate is 12 % annually
rate in quarterly = 12/4= 3%
We will apply the compound interest equation:
N=P( 1+r/n)^nt
N= ending number of cars serviced.
P= the number of cars serviced in 2012,
r = interest rate
n = the number of compoundings per year
t= total number of years.
Number of compoundings for t years = n*t = 4t
Initial number of cars serviced=920
The quarterly rate of growth = n=4
r = 3%
The growth rate = 1.03
Compound period multiplied by number of years = 920(1.03)^4t
Thus N = 920(1+0.03)^4t
N = number of cars serviced after t years...
Answer:
x ≈ 4.6
Step-by-step explanation:
Reference angle = 75°
Opposite side to reference angle = 17
Adjacent side = x
Applying TOA:
Tan 75 = opp/adj
Tan 75 = 17/x
Multiply both sides by x
x*Tan 75 = 17
Divide both sides by Tan 75
x = 17/Tan 75
x ≈ 4.6
If the troops hike 7km/2hr (read 7 kilometers per 2 hours), then to find how far they'd hike 4hr, just find multiply the top and bottom by 2. We do this because we found that to get from 2 to 4, we multiply by 2. And since this is a rate, we multiply the top by this value as well, so 7x2=14km
Complete Question
Evaluate the Fermi function for an energy kT above the Fermi energy. Find the temperature at which there is a 1% probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron.
Answer:
a
The Fermi function for the energy KT is 
b
The temperature is 
Step-by-step explanation:
From the question we are told that
The energy considered is 
Generally the Fermi function is mathematically represented as
![F(E_o) = \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }](https://tex.z-dn.net/?f=F%28E_o%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5BE_o%20-%20E_F%5D%7D%7BKT%7D%20%7D%20%2B%201%20%7D)
Here K is the Boltzmann constant with value 
is the Fermi energy
is the initial energy level which is mathematically represented as
So
![F(E_o) = \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}](https://tex.z-dn.net/?f=F%28E_o%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5B%5BE_F%20%2B%20KT%5D%20-%20E_F%5D%7D%7BKT%7D%20%7D%20%2B%201%7D)
=> 
=> 
=>
Generally the probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron is mathematically represented by the Fermi function as
![F(E_k) = \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 } = 0.01](https://tex.z-dn.net/?f=F%28E_k%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5BE_k%20-%20E_F%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
Here
is that energy level that is 0.5 ev above the Fermi energy 
=> ![F(E_k) = \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 } = 0.01](https://tex.z-dn.net/?f=F%28E_k%29%20%3D%20%20%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B%5B%5B0.50%20eV%20%2B%20E_F%5D%20-%20E_F%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
=> ![\frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 } = 0.01](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Be%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%2B%201%20%7D%20%20%3D%200.01)
=> ![1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01](https://tex.z-dn.net/?f=1%20%3D%200.01%20%2A%20e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%2B%200.01)
=> ![0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }](https://tex.z-dn.net/?f=0.99%20%3D%200.01%20%2A%20e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D)
=> ![e^{\frac{0.50 eV ]}{KT_k} } = 99](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B0.50%20eV%20%5D%7D%7BKT_k%7D%20%7D%20%20%3D%2099)
Taking natural log of both sides
=> 
=> 
Note eV is electron volt and the equivalence in Joule is 
So
=>
Answer:
9y + 24
Step-by-step explanation:
2(3y + 6) - 3(-4 - y)
= 6y + 12 - 3(-4 - y)
= 6y + 12 + 12 + 3y
= 9y + 12 + 12
= 9y + 24
