That's correct.
You can do it either by direct counting of permutations, or by conditional probability (Bayes' thm).
If by counting, let n1 be the total number of permutation containing exactly one jack (in the first 3 cards), n2 be the total number of permutations containing exactly 2 jacks...
Be careful with the wording in part b. "Considering that in the first 3 cards there is a jack"
The results are different depending upon whether you take that to mean "exactly one jack", or whether you take it to mean "one or more jacks". Personally I what have thought it meant one or more...
No, this is the part that you are not understanding. You are never finding the exact value of the integral, because you never have an exact function to integrate in the first place. Numerical integrals like Simpson rule DO find the exact value of an integral, the integrand being the parabola...
The whole thing is inherently an approximation, from the measurements themselves down to the accuracy to which the spines fit the actual profile. Even if you do "exact" (algebraic) integration of the resulting spline functions, it's still an approximation as the spline itself is an...
Ok that makes sense to do it that way then. Normally the easiest way to measure the volume with a physical bottle you have in hand is to pour the contents into measuring jug or other suitable container. Or if measuring the outer volume of a complex shape to submerge it and measure the volume...
Ok that's good to know, it's an actual bottle. Are you trying to find the "outer" volume or are you accounting for the wall thickness?
Yes by arbitrary positions I mean can you measure the radius at arbitrary points along the axis. From your description it appears that you can.
Was it asked...
What form is the data that you have about the bottle. Is it given as a scaled drawing, or is it given as a discrete set of points? (or perhaps it's an actual physical bottle you have?) I'm wondering if you can you sample the radii at arbitrary locations or not.
It would help if you could you...
If you're just using measured points then you're usually best off to use a numerical integrator like Simpson's rule or a Gauss quadrature. Essentially these methods both fit a polynomial to the points and calculate the area/volume in the one hit.
For the case of Simpson's rule the polynomial...
Just sketching out a proof based on anuttarasammyak's insights.
\frac{1}{k! \, n!} + \frac{-1}{(k+1)! \, (n-1)!} + \frac{1}{(k+2)! \, (n-2)!} \cdots \frac{(-1)^n}{(k+n)! \, (0)!} = \frac{1}{(k-1)! \, n! \, (k+n)}
Multiplying both sides by (k+n)! puts the proposed identity into the following...
Thanks all. Yes I did try induction first up, but I didn't get anywhere with it.
That's a great approach anuttarasammyak, looks very promising.
And using the combinations property that,
_nC_r = \, _{n-1}C_r + \, _{n-1}C_{r-1}
with the end condition that _nC_0 = \, _{n-1}C_0, I'm pretty...
This alternating series indentity with ascending and descending reciprocal factorials has me stumped.
\frac{1}{k! \, n!} + \frac{-1}{(k+1)! \, (n-1)!} + \frac{1}{(k+2)! \, (n-2)!} \cdots \frac{(-1)^n}{(k+n)! \, (0)!} = \frac{1}{(k-1)! \, n! \, (k+n)}
Or more compactly,
\sum_{r=0}^{n} (...
I can see some benefits for early students in terms connecting it to the parabola axis of symmetry (-b/2a), and also some good practice with difference of two squares expansion/factorization. Also could be useful in introducing students to the concept of substitution of a variable in an...
BTW, the method the book uses is the standard way to solve DEs of the form \frac{d^2 y}{dx^2} = f(y)
Define the first derivative as a new variable, eg v = \frac{dy}{dx}.
Now in terms of v the original DE can be written as:
\frac{dv}{dx} = f(y)
Applying the chain rule you get:
\frac{dv}{dy} \...
Write your second derivative as \frac{d}{dx} \left( \frac{dV}{dx} \right) and it should be clear that when you try to separate it using your method, what you actually get is:
\sqrt{V} \, d (\frac{dV}{dx}) = \beta \, dx