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
Given ---›
y = -3
&
y = x - 0.8
So,
=> x = y + 0.8
=> x = -3 + 0.8
=> x = -2.2
Therefore, the best approximation for the solution to this system of equations is = (–2.2, –3)
Answer:
x = 10
Step-by-step explanation:
You can try the answers to see which works. (The first one does.)
Or, you can solve for the variable:
Divide by 75
... (1/5)^(x/5) = 3/75 = 1/25
Recognize that 25 = 5^2, so ...
... (1/5)^(x/5) = (1/5)^2
Equating exponents, you have
... x/5 = 2
... x = 10 . . . . . multiply by 5
_____
You can also start by taking logarithms:
... log(75) +(x/5)log(1/5) = log(3)
... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log
... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression
... x = 10 . . . . . multiply by 5
_____
"Equating exponents" is essentially the same as taking logarithms.
Since this is a right triangle, use the Pythagorean Theorem to find c.
Pythagorean Theorem ⇒ a² + b² = c², where a and b are the legs and c is the hypotenuse.
Plug in 10 and 24 for the legs and c for the hypotenuse.
10² + 24² = c²
100 + 576 = 676
√676 = 26
c = 26
To multiply you have to distribute. 3(h-5) will be 3h-3(5) which is 3h-15. So the answer is 3h-15
