Answer:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)
Step-by-step explanation:
Previous concepts
The half-life is defined "as the amount of time it takes a given quantity to decrease to half of its initial value. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not".
Solution to the problem
The half life model is given by the following expression:

Where A(t) represent the amount after t hours.
represent the initial amount
t the number of hours
h=2.6 hours the half life
And we want to estimate the % after 5.5 hours. On this case we can begin finding the amount after 5.5 hours like this:

Now in order to find the percentage relative to the initial amount w can use the definition of relative change like this:
% Remaining = 
We can take common factor
and we got:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining ![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)
Each class had 3.5 gallons of paint to paint with and 3 classes used 10.5 gallons of paint
Answer:
the coordinates are (2,-7)
Step-by-step explanation:
its this bacause if you add all the vertices then subtract them you will get this
Answer:
30 i think
Step-by-step explanation:
Answer:
Generally the barrier width is 
Step-by-step explanation:
From the question we are told that
The tunneling probability required is 
The barrier height is 
The electron energy is 
Generally the wave number is mathematically represented as
![k = \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }](https://tex.z-dn.net/?f=k%20%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7B2%20%2A%20m%20%5BV_o%20-%20E%5D%7D%7B%5C%3D%20h%5E2%7D%20%7D)
Here m is the mass of the electron with the value 
h is is know as h-bar and the value is 
So
![k = \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }](https://tex.z-dn.net/?f=k%20%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7B2%20%2A%209.11%20%2A10%5E%7B-31%20%7D%20%5B0.4%20-%200.04%5D%20%2A%201.6%2A10%5E%7B-19%7D%7D%7B%5B1.054%2A10%5E%7B-34%7D%5E2%5D%7D%20%7D)
=> 
Generally the tunneling probability is mathematically represented as
![T = 16 * \frac{E}{V_o } * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}](https://tex.z-dn.net/?f=T%20%20%3D%2016%20%2A%20%5Cfrac%7BE%7D%7BV_o%20%7D%20%20%2A%20%5B1%20-%20%5Cfrac%7BE%7D%7BV_o%7D%20%5D%20%2A%20e%5E%7B-2%20%2A%20k%20%2A%20a%7D)
So
![1.0 *10^{-5} = 16 * \frac{0.04}{0.4 } * [1 - \frac{0.04}{0.4} ] * e^{-2 * 3.0736 *10^{9} * a}](https://tex.z-dn.net/?f=1.0%20%2A10%5E%7B-5%7D%20%3D%2016%20%2A%20%5Cfrac%7B0.04%7D%7B0.4%20%7D%20%20%2A%20%5B1%20-%20%5Cfrac%7B0.04%7D%7B0.4%7D%20%5D%20%2A%20e%5E%7B-2%20%2A%203.0736%20%2A10%5E%7B9%7D%20%2A%20a%7D)
=> 
Taking natural log of both sides
![ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}](https://tex.z-dn.net/?f=ln%5B6.944%2A10%5E%7B-6%7D%5D%20%3D%20-2%20%2A%203.0736%20%2A10%5E%7B9%7D%20%2A%20a%7D)
=> 
=> 