Answer:
x=9
Step-by-step explanation:
Looking at the diagram, we see that four angles are formed by the intersection of two lines. The angles highlighted in blue and green are therefore, vertical angles. We know that vertical angles have equal angle measures meaning that in this case, the measure of the blue angle is equal to the measure of the green angle. Using this, we can now solve for x:
7x+49=2x+94
7x-2x=94-49
5x=45
x=45/5
x=9
<u>Answer:</u>
The correct answer option is C. Both A and B.
<u>Step-by-step explanation:</u>
We are given a figure of a rectangle which when transformed changes it length of two sides AB and DC to AB' and DC'.
By looking at the figure, we can conclude that the transformation has changed the lengths of some sides of the given figure as well as changed the area of the figure.
Therefore, the correct answer option is C.
Answer: 29
Step-by-step explanation:
3x2= 6
4x2=8
3x5=15
6+8+15= 29
hoped I helped you a bit! Good luck! :)
Answer:
The probability that on any given day the water supply is inadequate 
Step-by-step explanation:
Given
α = 2 and β = 3
As per Gamma distribution Function
Expanding the function and putting the given values, we get -
![[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991](https://tex.z-dn.net/?f=%5Btex%5D1%20-%20%5Cint%5Climits%5E9_0%20%7Bf%28x%3B2%2C3%29%7D%20%5C%2C%20dx%20%5C%5C1-%20%5Cint%5Climits%5E0_0%20%7B%5Cfrac%7B1%7D%7B9%7Dxe%5E%7B%5Cfrac%7B-x%7D%7B3%7D%20%7D%20%5C%2C%20dx%5C%5C%5C%5C%3D%201-%20%5Cfrac%7B1%7D%7B9%7D%20%5Bx%28-3e%5E%7B%5Cfrac%7B-x%7D%7B3%7D%7D%29%20-%5Cint%5Climits%20%7B%28-3e%5E%7B%5Cfrac%7B-x%7D%7B3%7D%7D%29%7D%20%5C%2C%20dx%5D%5E9_0%3D%201-%201%2F9%20%5Bx%28-3e%5E%7B%5Cfrac%7B-x%7D%7B3%7D%7D%29%20-9e%5E%7B%5Cfrac%7B-x%7D%7B3%7D%7D%29%7D%20%5C%2C%20dx%5D%5E9_0%5C%5C1-%28%28%5Cfrac%7B-1%7D%7B3%7D%20%2A9%2Ae%5E%28%7B%5Cfrac%7B-9%7D%7B3%7D%7D%29-%20e%5E%28%5Cfrac%7B-9%7D%7B3%7D%29%29-%20%28%28%5Cfrac%7B-1%7D%7B3%7D%20%2A0%2Ae%5E%28%5Cfrac%7B-9%7D%7B3%7D%29-e%5E%7B%5Cfrac%7B-0%7D%7B3%7D%7D%29%5C%5C1-%28%28-3e%5E%7B%5Cfrac%7B-9%7D%7B3%7D%20%7D-e%5E%7B%5Cfrac%7B-9%7D%7B3%7D%7D-%280-1%29%29%5C%5C1-%281-4e%5E%7B-3%7D%29%5C%5C1-%281-0.1991%29%5C%5C1-0.8009%5C%5C0.1991)
The probability that on any given day the water supply is inadequate
