Answer:
Step-by-step explanation:
1)
2) x/y=8
3) x*1/4 + x
4) 6w = 4L
5) 3x-2 = 5*y
Answer:
Would be in this order D, E, C, A
Step-by-step explanation:
1.) 3(5 - 2x) = -2(6 - 3x) - 10x
15 - 6x = -12 + 6x - 10x
-6x - 6x + 10x = -12 - 15
-2x = -27
x = -27/-2 = 13.5
x = 13.5
2.) 1/4x - 2/5x = 39
-3/20x = 39
x = (39 x 20)/-3 = -260
x = -260
3.) 115 = 20 + 2.5x
2.5x = 115 - 20 = 95
x = 95/2.5 = 38
x = 38
4.) 0.7x - 0.2 = 8.3x + 7.2
8.3x - 0.7x = -0.2 - 7.2 = -7.4
7.6x = -7.4
x = -7.4/7.6 = -37/38
x = -37/38
Answer:
Area =
square centimeter
If the circumference of the hub cap would have been smaller, then its radius would have been smaller and subsequently its area would also have been less.
Step-by-step explanation:
As we know the circumference of a circle is ![2\pi r](https://tex.z-dn.net/?f=2%5Cpi%20r)
The radius of the hub cap needs to be devised to determine the area
![= 2* 3.14*r](https://tex.z-dn.net/?f=%3D%202%2A%203.14%2Ar)
![r = \frac{83.96}{2*3.14}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B83.96%7D%7B2%2A3.14%7D)
centimeters
Area of the hub cap = ![\pi r^{2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D)
Substituting the devised value of r in the above equation, we get -
![3.14 * 13.37^2\\](https://tex.z-dn.net/?f=3.14%20%2A%2013.37%5E2%5C%5C)
square centimeter
If the circumference of the hub cap would have been smaller, then its radius would have been smaller and subsequently its area would also have been less.
Answer:
Step-by-step explanation:
a=0 and b=20 in the uniform density function
∴ the mean μ=
=
=10 and
the variance
=
=
.
The standard deviation is the square root of the variance, so ![\sigma =\sqrt{\frac{100}{3}}=\frac{10}{\sqrt{3}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%5Csqrt%7B%5Cfrac%7B100%7D%7B3%7D%7D%3D%5Cfrac%7B10%7D%7B%5Csqrt%7B3%7D%7D)
Having determined the mean and standard deviation of the uniform distribution, we can conclude that
follows a normal distribution with
and
.
The normal probability distribution is:
![f\left ( x \right )=\frac{1}{\sigma \sqrt{2\pi }}\varrho ^{-\frac{1}{2}}\ast \left ( \frac{x-\mu }{\sigma } \right )^{2}](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3D%5Cfrac%7B1%7D%7B%5Csigma%20%5Csqrt%7B2%5Cpi%20%7D%7D%5Cvarrho%20%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Cast%20%5Cleft%20%28%20%5Cfrac%7Bx-%5Cmu%20%7D%7B%5Csigma%20%7D%20%5Cright%20%29%5E%7B2%7D)
So, substituting
and
![f\left ( x \right )=\frac{\sqrt{3}}{\ 50\sqrt{2\pi }}\varrho ^{-\frac{1}{2}}\ast \left ( \frac{x-250 }{50\sqrt{3}} } \right )^{2}](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B%5C%2050%5Csqrt%7B2%5Cpi%20%7D%7D%5Cvarrho%20%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Cast%20%5Cleft%20%28%20%5Cfrac%7Bx-250%20%7D%7B50%5Csqrt%7B3%7D%7D%20%7D%20%5Cright%20%29%5E%7B2%7D)
Having approximated sum
, we move on to the standardized sum
, which is the same
as only with μ=0 and σ=1. This means the probability distribution
is the standard normal distribution, which is:
![f\left ( x \right )=\frac{1}{\sigma \sqrt{2\pi }}\varrho ^{\frac{-1}{2}x^{2}}](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3D%5Cfrac%7B1%7D%7B%5Csigma%20%5Csqrt%7B2%5Cpi%20%7D%7D%5Cvarrho%20%5E%7B%5Cfrac%7B-1%7D%7B2%7Dx%5E%7B2%7D%7D)