Answer:
The 90% confidence interval for population mean is 
Step-by-step explanation:
From the question we are told that
The sample mean is 
The confidence level is 
The sample size is 
The standard deviation
Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as


Next we obtain the critical value of
from the standardized normal distribution table. The values is 
The reason we are obtaining critical values for
instead of that of
is because
represents the area under the normal curve where the confidence level 1 -
(90%) did not cover which include both the left and right tail while
is just considering the area of one tail which is what we required calculate the margin of error
Generally the margin of error is mathematically evaluated as

substituting values


The 90% confidence level interval is mathematically represented as

substituting values



Y = 2x + 0.5
or
y = 2x + 1/2
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)
=> 
=> 
La forma estándar de una ecuación de un círculo es% 28x-h% 29% 5E2 +% 2B +% 28y-k% 29% 5E2 + = + r% 5E2
donde Pt (h, k) es el centro y r es el radio
x ^ 2 + y ^ 2-10x + 4y + 4 = 0
completando los cuadrados para ponerlos en la forma estándar
x ^ 2 -10x + y ^ 2 + 4y + 4 = 0
(x-5) ^ 2-25 + (y +2) ^ 2-4 + 4 = 0
(x-5) ^ 2 + (y + 2) ^ 2 = 25
El centro es (5, -2) radio = 5
I probably did this wrong but i think its 21780.....cause i timed an acre by 1/2 to get that answer <span />