Let the number be Y .
Y is seven
less than 5 times another (let that other be X)
Y=7-5X
sum of both Y and X minus 4 gives one
Y+X-4=1
now solve both equations .
<h3>
Answer: 16/33</h3>
It's in p/q form where p = 16 and q = 33.
=======================================================
Work Shown:
x = 0.484848.....
100x = 48.4848.....
I multiplied both sides by 100 to move the decimal over 2 spots. Both decimal values for x and 100x have an infinite string of "48"s repeated after the decimal point. When we subtract, those infinite strings will cancel out
100x - x = 99x
48.4848..... - 0.484848..... = 48
So after subtracting straight down, we have the new equation 99x = 48 which solves to x = 48/99
Divide both parts by the GCF 3 to fully reduce
48/3 = 16
99/3 = 33
Therefore, x = 48/99 = 16/33 = 0.484848...
I recommend using a calculator to confirm that 16/33 = 0.484848...
Side note: your calculator may round the last digit, but this is of course rounding error
9514 1404 393
Answer:
(b) 4x- 3y = 6
Step-by-step explanation:
Standard form is ...
ax +by = c
where a, b, c are mutually prime integers and a > 0. (If a=0, then b>0.)
__
A: slope-intercept form with variables on both sides of the equal sign.
B: standard form
C: coefficients aren't integers
D: coefficients aren't integers; leading coefficient is negative.
Guatemalans would use milliliters because they use the metric system which is common throughout Latin America and which is basically easier to use than the Imperial system because it is all divisible by 10 whether it be millitres,litres,meters, kilometers etc.
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
