Answer:
Solution:
the boat's speed in still water: x
The speed with current: (x+3)
The speed against current: (x-3)
The equation is
3(x+3) = 4(x-3)
x = 21 mph
The distance = 3*(21+3)= 72 miles
I choose (A)
Step-by-step explanation:
Y=-7 use distributive property add like terms
Answer:
C. (x, y) → (x, -y)
Step-by-step explanation:
The algebraic representation that correctly describes a reflection over the x-axis is (x, y) → (x, -y)
Here's an example to show understanding
Plot A is (4,6) and the reflection over the x-axis would be (4,-6)
Answer:
1) 5
2) 5
Step-by-step explanation:
Data provided in the question:
(3²⁷)(5¹⁰)(z) = (5⁸)(9¹⁴)(
)
Now,
on simplifying the above equation
⇒ (3²⁷)(5¹⁰)(z) = (5⁸)((3²)¹⁴)(
)
or
⇒ (3²⁷)(5¹⁰)(z) = (5⁸)(3²⁸)(
)
or
⇒ 
or
⇒
or
⇒
we can say
x = 5, y = 2 and, z = 3
Now,
(1) y is prime
since, 2 is a prime number,
we can have
x = 5
2) x is prime
since 5 is also a prime number
therefore,
x = 5
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 