GingerAle

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UsernameGingerAle
Score2511
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Questions 4
Answers 740

 #2
avatar+2511 
+2

The malfunction appears to have started sometime after 2022-02-16 09:22:54 GMT

The most recent post to have more than one view is https://web2.0calc.com/questions/fraction-sequence, which has seven (7) views. This question post was answered by Alan 21 minutes after the initial post. This is extremely low for any post with this age, and is especially so for a post answered by Alan.  (This is probably unrelated, but it’s interesting to note that three (3) sequential prior post threads are hidden from public view).

 

When Alan posts an answer on a thread, the view count rate for that thread will climb about *.* times the average rate for all threads. This measure is calculated when the answer post appears at the top of the “Latest” page of “All Answers” and ends when it falls-off-the-bottom (with post per gage set to 25). At this point, the view count rate for a thread where Alan answers jumps to *.* times the average, then climbs slightly as the thread ages.  It’s quite funny to note Alan’s view count rate is now seven (7) times the average rate. This is, by far, the highest rate I’ve measured for anyone. An outlier for sure! LOL

 

Alan’s rates (and measured members) were higher a year ago. Not because Alan’s view counts were substantially higher –they’ve remained fairly steady, but relative to the overall average they climbed as the total view count dropped because the questions posted on the forum became more Brain-Dead.  Very few want to read third-grade math questions, asked by remediated high school students. When Alan answers a question it’s usually advanced enough to glean the interest of the more advanced students and mathematicians.

 

Another reason for the change is the link to the answers index is no longer available for users who are not logged on. Many or most do not realize the All Answers page is accessible by typing in the URL.   That’s not surprising! ...Not to me, at least. W T F is a URL?

 

 

GA

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Feb 18, 2022
 #2
avatar+2511 
+2

I was curious about the same thing: W T F

https://web2.0calc.com/questions/please-help_76830#r2

 

Aside from different power and distance measurements, this question uses a parenthetical (BS) along with base station. However, I think the (BS) is (now) included by the author as a cryptic indication that the physics in the question assumes facts not in evidence: That is, the question is Bull Shit, even though the equation is correct as a baseline for EM waves reflected back to a transmitter, or as a baseline to correct errors in time-domain reflectometry measurements. 

 

GA

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Feb 14, 2022
 #6
avatar+2511 
+3

Hi Melody,

The graphic below depicts the \(2^3 = 8\) arrangements of (statistical) success, where each person receives a red card.  I used this formula here: https://web2.0calc.com/questions/pleaze-help#r9 where there are three sets of three colors.

 

\(\hspace {.15em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.02em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ Y & Y & Y \\ \end{array} } \right] \text {First arrangement of (R} \scriptsize{s} \normalsize{) } {\; \mathrm and \; \mathrm (Y} \scriptsize{s} \normalsize{)} \\ \left[ {\begin{array}{ccc} R & R & Y \\ Y & Y & R \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & Y & R \\ Y & R &Y \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & Y & Y \\ Y & R & R \\ \end{array} } \right] \text {Fourth arrangement} \\ \left[ {\begin{array}{ccc} Y & R & R \\ R & Y & Y \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{ccc} Y & R & Y \\ R & Y & R \\ \end{array} } \right] \text {Sixth arrangement} \\ \left[ {\begin{array}{ccc} Y & Y & R \\ R & R & Y \\ \end{array} } \right] \text {Seventh arrangement} \\ \left[ {\begin{array}{ccc} Y & Y & Y \\ R & R & R \\ \end{array} } \right] \text {Eighth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .}\\ \, \\ \small \text {For all other arrangements: }\\ \small \text {one person has zero red (0) cards, one person has one (1) red card, and one person has two (2) red cards.} \\ \)

 

GA

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Feb 14, 2022
 #4
avatar+2511 
+3

Hi Melody,

You correctly solved this problem here: https://web2.0calc.com/questions/probability-question_88#r3

 

Confirmation: \(\large (2^3) / (nCr(6,3) =  \dfrac {2}{5} = 40\%\)

 

 

GA

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Feb 13, 2022
 #7
avatar+2511 
+1

 

Here is a formal solution, using Euler's totient function –calculated from non-prime numbers. 

There are often (not always) solutions to simultaneous modular equations with non-co-prime modulo numbers, when one of the co-primes modulos is equivalent to zero.

 

The first product of zero (0) is included as a presentation continuity formality.

 

 

\(\begin{array}{rcll} n &\equiv& {\color{red}0} \pmod {{\color{green}8}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}9}} \\ n &\equiv& {\color{red}4} \pmod {{\color{green}10}} \\ \text{Set } m &=& 8\cdot 9\cdot 10 = 720 \\ \end{array}\)

 

\({ \begin{array}{l} n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}9 \cdot 10) }^{\varphi({\color{green}8}) -1 } \pmod {{\color{green}8}} ] }_{=\text{modulo inverse }(9\cdot 10) \mod 8 } }_{=(9\cdot 10)^{4-1} \mod {8}} }_{=(9\cdot 10)^{3} \mod {8}} }_{=(90\pmod{8})^{3} \mod {8}} }_{=(2)^{3} \mod {8}} }_{= 0} + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 10) }^{\varphi({\color{green}9}) -1} \pmod {{\color{green}9}} ] }_{=\text{modulo inverse } (8\cdot 10) \mod {9}} }_{=(8\cdot 10)^{6-1} \mod {9}} }_{=(8\cdot 10)^{5} \mod {9}} }_{=(80\pmod{9})^{5} \mod {9}} }_{=(8)^{5} \mod {9}} }_{=8} + {\color{red}4} \cdot {\color{green}8\cdot 9} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}8\cdot 9) }^{\varphi({\color{green}10}) -1 } \pmod {{\color{green}10}} ] }_{=\text{modulo inverse } (8\cdot 9) \mod 10 } }_{=(8\cdot 9)^{4-1} \mod { 10}} }_{=(8\cdot 9)^{3} \mod {10}} }_{=(72\pmod{10})^{3} \mod {10}} }_{=(2)^{3} \mod {10}} }_{=8}\\ \\ n = {\color{red}0} \cdot {\color{green}9\cdot 10} \cdot [0] + {\color{red}1} \cdot {\color{green}8\cdot 10} \cdot [8] + {\color{red}4} \cdot {\color{green}8\cdot 9} \cdot [8] \\ n = 0+ 640 + 2304 \\ n = 2944 \\ n \pmod {m} \rightarrow 2944 \pmod {720} \equiv 64 \\ \\ n = 64 + k\cdot 360\qquad k \in Z\\ \hspace {10em} \mathbf{n_{min}} \mathbf{=} \mathbf{64} \end{array} } \)

 

 

GA

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Feb 10, 2022