GingerAle

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Имя пользователяGingerAle
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 #2
avatar+2511 
-6

About every two years, usually during an odd year, some idiot posts a simple expression with a wrong answer for its simplification and resolution to a numerical result. The same idiot also claims the calculator does not solve it correctly because it fails to follow the order of operations [PEMDAS or BIDMAS] correctly.

 

This year, you are the idiot!

 

 The expression, as presented, equals (25) not one (1).

 

If by “distribution rules” you mean Arithmetic Order of Operation Hierarchy, then this calculator follows all (except for one*) of them, including the exceptions for variables usage, where Implicit multiplication of variables takes precedents over division – a formally adopted exception, dating back to the 1968 mathematical society convention that included updates on  mathematical order of operations hierarchy.

 

As for your brain-dead expression, the calculator correctly resolves this to (25).

20/4(3+2) = 25  | This is correct for Arithmetic Order of Operation Hierarchy

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Note:  by using a variable, then this resolves to one (1). This makes use of the exceptions for implicitly multiplied variables before division (described above).

x=4; 20/(x)(3+2) = 1  Though the parenthetical operation of (3+2) is not a variable it’s treated as one because of the variable preceding it and the implicit multiplication.

 

Also, by placing parenthesis around the (4), the calculator will use the variable-exception rule.

20/(4)(3+2) = 1  Again, note that multiplication is implied between the parenthetical operations.

[It’s also notable that when math is resolved or solved electronically, parenthetical operations are treated as variables in the registers.]

If the multiplication is made explicit with a * then:

20/(4)*(3+2) = 25 ... the operations revert back to standard PEMDAS.

 

This matches the answer you claim is correct, but ...

 

Your expression does not have a variable, so the exception for implicitly multiplied variables before division does not apply. This calculator is smarter than you are!

 

Additional notes:

*This calculator’s exception for Arithmetic Order of Operation Hierarchy occurs for the Stacked Power convention:

 

Formal operation: Stacked powers (aka: Power Towers) are (exponentially) multiplied from the right to left (from the top down), and the resultant product becomes the EXPONENT to the base number.

 

For this calculator, the Arithmetic Order of Operation Hierarchy is not followed for Stacked Powers. The web2.0calc calculator resolves this from the left to right or ascending order, where the resultant product becomes the BASE of the next exponent. This operation is clearly defined in the interpretation space, above the calculator. 

 

Generally, power-towers are used in advanced, theoretical mathematics.  For the majority of occasions where the construction of a power-tower occurs in an equation for the physical sciences, the ascending method is almost always used for its resolution (the descending method is rarely used). This is probably why Herr Mossow elected to code the calculator with the ascending method as the default operation.   This calculator, which is actually a computational engine, is orientated toward the physical sciences, which is why it’s referenced as a Scientific Calculator. 

 

Related post: https://web2.0calc.com/questions/8-2-2-2

 

GA

--. .-

18 янв. 2023 г.
 #4
avatar+2511 
-6

TooEasy, your answer is not only WRONG, your solution is ATROCIOUS!

Posting alibi disclaimers to negate and shirk responsibility for mathematical SLOP

are hallmarks for mathematicians who are worth less than nothing.

 

 

Solution:  My solution is Wrong and it’s SLOP...

 

Description:

Choose one (1) of the seven (7) numbers, then choose three (3) of the four (4) colors for these numbers.

Then choose one (1) of six (6) numbers and then choose one (1) of the four (4) colors for this number.

Then choose one (1) of six (6) numbers (again), and then choose one (1) of the four (4) colors for this number (again).

 

Then divide this by the \(\dbinom {27}{5}\)ways to choose a set of five (5) card from a population of 27 cards. 

 

Descriptive Math:

 

\(\text {Ignore this slop}\\ \dfrac { \dbinom {7}{1} \dbinom {4}{3} \dbinom {6}{1} \dbinom {4}{1} \dbinom {6}{1} \dbinom {4}{1}} {\dbinom {27}{5}} = \underbrace {\left (\large \dfrac {896}{4485}\right)}_{\text {exact probability}} \left ( \approx 19.98\% \right )\)

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Even if I corrected the divisor, it’s still SLOP...

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To solve at this level requires hypergeometric distributive counting. (Tiggsy’s solution)

OR

Case analysis of binomial distribution. (Below)

 

\(\huge \rho(x) \normalsize = \dfrac { \left [\dbinom {7}{1} \dbinom {4}{3} \right] * \left[ \left (\dbinom {6}{2} \dbinom {4}{1}\dbinom {4}{1} \right ) + \left(\dbinom {6}{1} \dbinom {4}{2}\right) \right]} {\dbinom {28}{5}} = \underbrace {\left (\large \dfrac {46}{585}\right)}_{\text {exact probability}} \left ( \approx 7.86\% \right )\\ \)

Expansion, Dissection, and Description:

 

nCr(7,1)*nCr(4, 3) * nCr(6,2)*nCr(4, 1)*nCr(4, 1) =  6720  |Counts of triples with non-pairs.

 

Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose two (2) of the six (6) remaining numbers; choose one (1) of four colors for each of the two (2) numbers.  

 

nCr(7,1)*nCr(4, 3) * (nCr(6,1)*nCr(4, 2) = 1008 |Counts of triples and only pairs.

 

Choose one (1) of seven (7) numbers; choose three (3) of the four (4) colors; choose one (1) of the six (6) remaining numbers; choose two (2) of the four (4) colors –making a pair of numbers. 

 

Add these counts: 6720 + 1008 = 7728

 

Divide: 7728/ nCr(28,5) =  46/585  = 7.86%

 

GA

--. .- 

2 янв. 2023 г.
 #15
avatar+2511 
-5

 

Solution for Q1:

(Because a member can abstain, this is counted as a vote for no candidate.)

 

Case one (1): All candidates receive at least one vote, including the no candidate, which means there is at least one abstention.

                     Partitions of 50 with a size of 5 = 2611

 

Case one (2): One candidate receives zero (0) votes; all others receive at least one vote.

                       Partitions of 50 with a size of 4 = 920

 

Case one (3): Two candidates receive zero (0) votes.

                      Partitions of 50 with a size of 3 = 208

 

Case one (4): Three candidates receive zero (0) votes.

                      Partitions of 50 with a size of 2 = 25

 

Case one (5): Four candidates receive zero (0) votes. This means everyone abstains, i.e. everyone votes for no candidate

                      Partitions of 50 with a size of 1 = 1

 

 

Total distribution of votes: 2611 + 920 + 208 + 25 + 1 = 3765

------------------

 

Solution for Q2:

 

(This is same as question 1, except the members cannot abstain.)

 

Case one (1): All candidates receive at least one vote

                      Partitions of 50 with a size of 4 = 920

 

 

Case one (2): One candidate receives zero (0) votes; all others receive at least one vote.

                       Partitions of 50 with a size of 3 = 208

 

 

Case one (3): Two candidates receive zero (0) votes.

                      Partitions of 50 with a size of 2 = 25

 

Case one (4): Three candidates receive zero (0) votes. One candidate receives all the votes.

                      Partitions of 50 with a size of 1 = 1

 

 

Total distribution of votes:  920 + 208 + 25 + 1 = 1154

 

 

 

GA

--. .- 

7 нояб. 2022 г.