First of all, I'm going to assume that we have a concave down parabola, because the stream of water is subjected to gravity.
If we need the vertex to be at
, the equation will contain a
term.
If we start with
we have a parabola, concave down, with vertex at
and a maximum of 0.
So, if we add 7, we will translate the function vertically up 7 units, so that the new maximum will be 
We have

Now we only have to fix the fact that this parabola doesn't land at
, because our parabola is too "narrow". We can work on that by multiplying the squared parenthesis by a certain coefficient: we want

such that:
Plugging these values gets us

As you can see in the attached figure, the parabola we get satisfies all the requests.
Answer:
0
Step-by-step explanation:
given that we roll a fair die repeatedly until we see the number four appear and then we stop.
the number 4 can appear either in I throw, or II throw or .... indefinitely
So X = the no of throws can be from 1 to infinity
This is a discrete distribution countable.
Sample space= {1,2,.....}
b) Prob ( 4 never appears) = Prob (any other number appears in all throws)
= 
where n is the number of throws
As n tends to infinity, this becomes 0 because 5/6 is less than 1.
Hence this probability is approximately 0
Or definitely 4 will appear atleast once.
Answer:
With j = number of jars: 2,578 / 6 = j
Step-by-step explanation:
[ - 6 * x^2 * y^8 + 12* x * y^3 - 36 * x * y^2 ] / [6*x*y^2] =
[-6x^2 y^8 ] / [6xy^2] + [12x y^3] / [6xy^2] - [36xy^2] / [6xy^2] =
- xy^6 + 2y - 6
Answer: - xy^6 + 2y - 6