член: GingerAle

Actually, you are ...in a metaphorical sense. You’ve demonstrated such skills on more than one occasion. I think most teachers need above average skills in mind reading to be a good teacher. Beyond basic mind-reading is the quintessential skill of filling in the thoughts for thought processes that are missing essential information in the student’s mind.

For teaching younger students, this is a common and relatively easy requirement because younger students are usually just missing information. The difficulty is in persuading the student to use his/her mind to its full capacity.

For the older student, however, the requirements are more arduous because wrong and errant information may be firmly entrenched. The student needs to be willing to excise errant thought process to make room for the correct processes.

There is a valid argument that to read a mind, there needs to be a mind to read. Mindlessness is a real phenomenon. It’s exceptionally difficult to comprehend and hence exceptionally difficult to treat.

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GingerAle23 сент. 2022 г.

+2440

-1

Solution: $\dfrac{5!}{2!} * \dbinom {6}{3}$

Description of process:

Removing the ‘C’s from the set gives (5!/2! =60) distinguishable arraignments for the remaining five (5) letters. The three ‘C’s can then occupy three (3) of the six (6) positions between the letters and at either end of the distinguishable arraignments

Graphic demonstrating positions: _A_U_R_A_Y_

Select the three (3) positions as $\dbinom {6}{3} = 20$

Total number of arrangements where no ‘C’s are adjacent to another ‘C’ $= (60 * 20) = 1200 $

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GingerAle15 авг. 2022 г.

+2440

-1

BuilderBoi, if you just used a kitchen scale, it would have resolved the ambiguity by indicating that N($3,456, $478) is in the form of $\mathbb{N}(\mu, \sigma)$ in this case there is no need for pot-laced candy or mushroom tea (though it might help ).

Recently, I saw a demonstration of very sophisticated kitchen scale at university. This scale connects to an AI-nerualnetwork. The scale’s sensors along with the AI identify the foods, caloric values, glycemic index, and other parameters. The AI also calculates the mean probable weight gain (with a standard deviation), based on the metabolic parameters of the consumer. It’s easy to program the parameters: the consumer need only to stand and then sit on the scale for 8 seconds, and then spit on a sensor. The demonstrated scale had a mass/weight limit of 450Kg –well above the weight of the largest tub-of-lard in attendance.

During the demonstration, one of the spectators plopped a bag of pot-laced candy on the scale. The AI correctly identified the contents and caloric value, but indicated an 85% probability of weight gain at a mean of (28.7) times the maximum value for the calorie count.

The engineer-tech, who was demonstrating the scale, queried the AI for an explanation. The AI extrapolated the stats for $\Delta9THC $ causing the munchies in a standard population; from this, a probability of weight gain is calculated beyond the caloric value of the weighed food.

It is amazing how innovative technology can create a kitchen scale that can construct and solve statistical problems. You should get one as soon as they are on the market.

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GingerAle10 авг. 2022 г.

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Hi Melody,

OMG! I saw two trees but missed the whole bloody forest.

[You’re in a flood, but I’m the one that’s all wet...]

Until I saw your graph, I had tunnel vision. I didn’t really do a (full) positional analysis. I only compared the positions of the two sets for each simultaneous sequence. The two diagonals are the only two that occur simultaneously in the 8! counts.

It’s Spooky Quantum Dumbness at a distance and I walked right into it, face first, because of quantum blindness.

Lancelot Link would throw banana peels and coconuts at me for that.

Very Cool! Thank you.

--. .-

GingerAle7 июл. 2022 г.

+2440

Hi Melody,

But I have to subtract all the ones that are in both every row and every column (becasue they have been added twice.)This number is 8!

You’ve subtracted (8!) from the total successful arrangements. That is a lot duplication that is NOT indicated in the positional analysis of the rooks. The only time there is a rook in every row and every column is when they are positioned diagonally across the board. Other than this, one or more rows and/or one more columns are unoccupied, but not unattacked.

As I described above, this diagonal arraignment occurs twice: once, descending from the top right to the bottom left, and again from the bottom left to the top right. These diagonal arraignments are the ONLY two arrangements that occur in both the vertical movement progression and the horizontal movement progression. There are no other duplicates, so where does the (8!) come from?

Appended to this post are images depicting the chessboard with eight rooks in horizontal and vertical formats, demonstrating their respective movements on the chessboard and points of interest.

If I am missing (8!-2) arrangements where these two formats intersect, then I have royally rooked myself. I would really like to see them. Can you graphically demonstrate where some of these duplications occur?

--. .-

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GingerAle4 июл. 2022 г.

+2440

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I agree with your interpretation, Ron:

It’s a quadratic expression divided by a linear expression.

$\large \frac{8x^2 - 4x} {4x – 2} \rightarrow 2x$ This is implied by the exception "that x isnt equal to 1/2" (sic).

However, because it is written in ASCII SLOP it needs parenthetical operators to confirm this, else it can (and should) be strictly interpreted as:

$\large 8x^2 – 4*\frac{x}{4}* x – 2 \rightarrow 7x^2 -2 $

Both the forum’s calculator and Wolfram Alpha interpret the expression using Formal ASCII Mathematical Hierarchy. There are good reasons why ASCII Mathematical Hierarchy was formalized by mathematicians, engineers, and computer scientists. ASCII was developed to communicate with computers, not to teach math to humans; Eventually its use evolved into communicating with humans ...and genetically enhanced Chimps...and along with it came ambiguity because the composers or the readers do not know the Mathematical Hierarchy.

Of course computers can be programmed to make assumptions –ignoring hierarchy; so there are online algebra solvers that will simplify or solve it as a quadratic expression divided by a linear expression, without the use of parenthetical operators.

--. .-

GingerAle2 июл. 2022 г.

+2440

Hello CPhill,

When calculating a statistic on indistinguishable elements or arraignments, they are (usually) treated as if they are distinguishable. Though not distinguishable as one from another, their appearance is proportional to their multiplicity, and this affects their appearance counts in statistical measures.

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GingerAle30 июн. 2022 г.

+2440

There is no casual solution to this problem; though the solution process is not ultra complex.

The use of the Inclusion-Exclusion principle will accelerate the process. Like usual, its application requires careful and (sometimes) laborious attention to avoid over-counting successes and failures.

The question resolves to a binary state of either an arrangement of eight (8) rooks occupying or attacking every square, or it does not. Analyzing the counts of unoccupied and un-attacked squares is not requires for this question.

(Permutations of the rooks among themselves are not necessary, as this will be done for both success and failure counts and will factor out.)

Setup:

Vertical format.

Starting with an 8 X 8 chessboard place eight rooks on the top row (row 1) and observe that every square is either occupied or attacked by a rook. Moving the rooks vertically (row-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically.

This same sequence is repeatable in a horizontal format.

Place the eight rooks in the left column (Column 1) observe that every square is either occupied or attacked by a rook. Moving the rooks horizontally (column-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically. HOWEVER, in this horizontal format, there are two (2) cases where the positions of the rooks identically match the positions in the vertical format. These two (2) cases occur when the rooks align diagonal positions across the board. This occurs twice: from the top-left to the bottom-right, and from the bottom-left to the top-right. These two (2) cases are exclusions and need to be subtracted from the total. So at this point, the subtotal for success counts are (8^8) + [(8^8) – (2)].

At this point it’s (apparently) discernable that for every square to be either occupied or attacked by a rook, every column or every row must have at least one rook. Removing all rooks from a column leaves some (not all) row squares on that column un-attacked. Like-wise, removing all rooks from a row leaves some (not all) column squares on that row un-attacked.

If the above is true, then these counts (8^8) + [(8^8) – (2)] are the only successes.

Probability of a random arrangement of eight (8) rooks to occupy or attack every square on a chessboard:

$\rho_{(s)} = \dfrac {(2*8^8)-2} {\binom {64}{8}} \approx 0.75809\%$

--. .-

GingerAle29 июн. 2022 г.

+2440

-1

Is this really your calculation? It looks like you copied it from somewhere.

If it is yours, then how did you formulate your calculation?

It is unlikely what you presented is the correct probablity.

Re-Present your calculation as an equation with the work product: delineate your calculations showing how the Inclusion-Exclusion principle applies to your calculation. Specifically, define the included set, the excluded set, and the union of these two sets.

If you do this, then it should be easy for someone to verify your solution, without deriving and solving this. (As an analogy, it’s easy to verify the primes of large numbers; it’s not so easy to find the primes in the first place.)

If you do decide to do this, use LaTex. Reading math in ASCII slop is often irritating.

--. .-

GingerAle29 июн. 2022 г.

+2440

-4

I think you have to weigh the paper, although a kitchen scale wouldn't be so precise. ...

BuilderBoi, are you nibbling on pot-laced candy or sipping on mushroom tea?

There’s nothing like a few psychotropic chemicals to help you in expanding your mind for thinking outside of the box.

Maybe the teacher used a previous question as a template and inadvertently left the kitchen scale in the current assignment. Or the teacher intentionally listed the use of a kitchen scale to see which students are paying attention.

Or... the teacher may have been nibbling and sipping...

Even without the kitchen scale blunder, this assignment is very poorly thought-out and poorly written with missing commas and superfluous commas, along with disjointed instructions connected with repetition; and meandering comments on measurement errors, which are not defined, even as a reference to a class lecture.

The teacher rambles on about Plainameters, referring to them as tools used prior to the digital era when, in fact, both mechanical and digital versions are still used. (Cabinet and furniture makers use them all the time.)

The students are welcome to search for and learn how this tool worked, but you don't need to include it in your write up. Why include what you don’t need to do in the main assignment? A better option is to include it as a footnote for optional research with a reference to its 0.2% relative error in the main instruction.

This is a math-based writing assignment from a teacher who does not know how to write.

I wonder if the teacher used the kitchen scale to distract the readers from the incompetent assignment presentation.

Everyone notices the kitchen scale instead of the crappy presentation and atrocious writing. The teacher may have thought of this while nibbling on pot-laced candy and sipping on mushroom tea.

--. .-

GingerAle21 июн. 2022 г.

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