I’m not sure what this answer is I will comment when I find it
Answer:
x=1+√334, x=1−√334
Step-by-step explanation:
Answer:
The ratio of the areas would be the ratio of the side lengths squared.
Say the ratio of the side lengths is 1:2, the ratio of the areas would be 1:4. If the ratio of the sides were 2:3, the ratio of the areas would be 4:9.
Hope this helps
For light: Inverse Square Law
<span>
The signal is weaken by the inverse square of the
difference of distances this means that one over the square of the
difference. </span>
<span>
</span>
Therefore, 70000LY / 141 LY =496.45 and that
squared =246462.602 times less signal, or 1/246462.602.
2 For a circular telescope like the
Arecibo radio telescope that equals Pi times the radius squared. Sensitivity
of the telescope depends on its area.<span>
The new telescope needs 246462.60 times that area to have
enough sensitivity </span>
<span>
<span>(150 x 150 x Pi) = 70 685 sq m area for the Arecibo telescope
and that times 246462.60
= 1.74212 x 10^10</span></span>
<span><span>
</span></span>
What is the diameter of a circle with that area?
<span>
<span>1. Divide the area by Pi = 5.548156 x 10^9
2. Take the square root of the result to find the radius = 74485.94293</span></span>
3. Multiply by 2 to find the diameter =
<span>148971.886</span>
Answer:
a) 
b) 0.0620
Step-by-step explanation:
We are given the following in the question:
Population mean,
= 6
Variance,
= 12
a) Value of 
We know that

Dividing the two equations, we get,

b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.
We can write the probability density function as:

We have to evaluate:
![P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620](https://tex.z-dn.net/?f=P%28x%20%3E12%29%5C%5C%5C%5C%3D%20%5Cdfrac%7B1%7D%7B16%7D%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_%7B12%7Df%28x%29dx%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B16%7D%5Cbigg%5B-2x%5E2e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D-2%5Cdisplaystyle%5Cint%20xe%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7Ddx%7D%5Cbigg%5D%5E%7B%5Cinfty%7D_%7B12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B8%7D%5Cbigg%5Bx%5E2e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%2B4xe%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%5Cbigg%5D%5E%7B%5Cinfty%7D_%7B12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B8%7D%5Cbigg%5B%28%5Cinfty%29%5E2e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%2B4%28%5Cinfty%29e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%20-%28%20%2812%29%5E2e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%2B4%2812%29e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%29%5Cbigg%5D%5C%5C%5C%5C%3D0.0620)
0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.