GingerAle

Greetings Asinus,

Question three (3) is missing. I included it below. 

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General definitions:

\(\text {The specific heat capacity of water is } \mathrm {4.1868 J \cdot g^{-1} \cdot C°^{-1}} \hspace{1em} | \hspace{1em } \small \text {Joule per gram per degree Celsius}\\ \)

1(a): Costs in general

\(\text {1800watts = 1.8kW}\\ \text {1.8kW * 6ct/kWh = 10.8ct/hour of operation }\\ \text {Operating this system for 15 minutes will cost 10.8/4 = 2.7ct }\\\)

1(b): Indicated values

\(\text {Quantity of water: 1 L = 1 Kg }\\ \text {Temperature change: Delta °C = 70°C | (85-15) }\\ \text {Power input to system: 1800 Watts | Watts/hour = 6480000J/hour }\\ \text {System Time for temperature rise: 15 minutes. }\\ \text {System operates for 15 minutes: cost of operation } \fbox {2.7ct} \hspace{1em}|\hspace{1em}\small \text {Same as costs in general }\\\)

2. Efficiency of the system:

\(\text {Input Energy required for 70} \hspace{-.1em} \mathrm {°C} \text { temperature rise for 1Kg water: }\\ E_J = 70\hspace{-.1em} \mathrm {°C} * \mathrm{1000g * 4.1868 J \cdot g^{-1} \cdot ° \hspace{-.1em} \mathrm {C^{-1}}}  = \mathrm {293076J }\\ \text {@ 100% efficiency: }\\ kWh = \mathrm {\frac{293076J} {3600000J/kWh} = 0.08141 \hspace{.15em} \small kWh} \hspace{1em} | \small \text { Required energy to raise temperature by 70°C for 1 liter of water } \\ \text {Time to transfer energy to water @ 1800 Watts: } \frac {293076J}{1800J/s } = \mathrm{162.82s} = \text{2 minutes 43 seconds.} \\ \text {Cost of energy:  0.08141kWh * 6ct/kWh \aprox 0.49ct }\\ \text {Actual time to heat water: 15 minutes. }\\ \text {Energy used by system to heat water: }  \frac{1.8kWh}{4} = 0.45kWh\\ \text {Efficiency of system: } \frac {0.08141kWh}{0.45kWh} \approx \fbox{18.09%}\\ \)

3. The general cost of warming the cold tits of a 60kg-witch to those of a hot babe.

\(\text {Assume: 39°C temperature rise, standard human physiology, non-frozen tits, }\\ \text{ and core body temperature approximates temperature.of tits }\\ \text{Specific heat of human body: }  \mathrm {3.470 \hspace{.1em}J \cdot g^{-1} \cdot ° \hspace{-.1em} C^{-1} }\\ \text {Input Energy required for 39°C temperature rise for 60kg human witch: }\\ \mathrm {E_J = 39°C * 60000g * 3.470 \hspace {.1em} J \cdot g^{-1} \cdot ° \hspace{-.1em} C^{-1} = 8119800 \hspace{.1em} J}\\ \mathrm{kWh = \frac{8119800J} {3600000J/kWh} \approx 2.26 kWh  }\\ \text{Time to transfer energy @1800 Watts: } \mathrm{\frac {8119800J} {1800} = 4511s}\\ \approx 1.25 \text{ hours @100% efficiency and 6.9 hours @ 18.09% efficiency. }\\ \text{Cost: 75ct at 18.09% efficiency}\\ \)

Notes: results may vary depending on the type of witch, but the tits will be much warmer. 

GA

Guest’s solution is correct. Mine is NOT.

My partial solution (copied from my archive of solved equations) is based on similar equations with a different set of congruencies (mod 13).

\(\begin{align*} 2ab+bc+ca&\equiv abc \pmod{13}\\ ab+2bc+ca&\equiv 3abc\pmod{13}\\ ab+bc+2ca&\equiv 5abc\pmod {13} \end{align*} \)

I failed to modify the congruencies to match the question.  

However, this solution method will work for the equations in the OP’s question.

GA

--. .-

Solution:

\(\text{Modular arithmetic theorem: }\\ \text {If (a, b, c) are positive integers and} \pmod{m} \text { is relatively prime, then there exists }\\ (a^{-1}, b^{-1}, c^{-1}) \text { such that } \left(a*a^{-1} \equiv b*b^{-1} \equiv c*c^{-1} \pmod{m} \right), \\\text{ with a, b, c, relatively prime within the ring of modulo (m); i.e. } 1\leq [a’, b’, c’] < 13. \)

\(\begin{array}{|rccc|} \hline a^{-1}b^{-1}c^{-1} (2ab + bc + ca) \equiv 1 \pmod{13}\\ a^{-1}b^{-1}c^{-1} (ab + 2bc + ca) \equiv 3 \pmod{13} \\ a^{-1}b^{-1}c^{-1} (ab + bc + 2ca) \equiv 5 \pmod{13}\\ \hline\end{array} \Rightarrow \begin{array}{|rccc|} \hline 2c^{-1} + a^{-1} + b^{-1} \equiv 1 \pmod{13}\\ c^{-1} + 2a^{-1} + b^{-1} \equiv 3 \pmod{13}\\ c^{-1} + a^{-1} + 2b^{-1} \equiv 5 \pmod{13}\\ \hline\end{array} \)

\(\text {add equations: }\\ 4c^{-1} + 4a^{-1} + 4b^{-1} \equiv 9 \pmod{13}\\ 4(c^{-1} + a^{-1} + b^{-1}) \equiv 9 \pmod{13}\\ \)

\(\text {Note that} -4\pmod{13}\equiv 9, \text{ so } \left (a^{-1} + b^{-1} + c^{-1}\right) \equiv -1\\ \text {and } -1 \equiv 12 \pmod{13},\text { so search for three numbers where } 1 \leq a,b,c < 12.\\ \text {using the above inverse equations to find which value belongs to each variable. }\\ \text{Then take the Modular Multiplicative Inverse of these numbers, sum them to (S) and find } \mathrm S\pmod{13}. \\ \text {if you do this correctly, you will find the answer (remainder) is } \boxed {2}\\ \)

GA

--. .-

Solution:

You can derive the initial velocity from the average velocity via the traveled distance where the breaking begins to the stop sign:

\(\text {The average velocity is } \dfrac{3.3 \tiny \;meter \hspace{-.1em}s} {2.4 \;s} = 1.375 \tiny \dfrac {m}{s}, \normalsize \text { then multiply by (2) for the initial velocity } (2 *1.375) = 2.75\tiny \dfrac {m}{s} \\ \text {Then }\\ a = \dfrac{v - u}{t} \hspace{1em}| \hspace{1em} \small \text {a = acceleration, v = final velocity, u = initial velocity, t = time} \\ \dfrac{0 - 2.75 \tiny m/s}{2.4s} = -1.5625 \; ms^{-2} \Leftarrow \text {negative acceleration} \)

GA

--. .-

Note: “Help” is spelled with an “e” not an “a” so, reserve the baby-talk for babies. Most babies do not know how to read....

I must not understand the question, because this could go on forever and you couldn't write down all of them if you keep writing until the end of time. 

Hi Ron,

You understand the question correctly. Your answer is correct. 

The formal solution is (q < 0)

It’s relationship to (x) is:  q = -x (x + 3)

...And most importantly, you remembered to "mind your peas and queues" 

GA

--. .-

...  I want to calculate the cost of giving out five thousand bonuses worth 100 each, i.e 500000 total with a turnover requirement of 36 times the bonus.

Well... obviously the cost (to the house) would be 500,000, if all the gamblers meet the turnover requirements.

For this case,

EV = 100 – (3600 * 0.03) = -8

This means house can expect to pay out a net (500,000 – (5000 * 8)) = 460,000, if all 5000 gamblers meet the turnover requirements.

This would be an incentive for new and long-gone gamblers to visit and return. This type of promotion would be paid for via an advertizing and promotion account. It would not be for the casino to realize a short term profit.

My original formula 100*(0.97^36) gives an EV of 33.4 which seems to be in the right ballpark for a 100 bonus?

I’m unfamiliar with this ballpark, and this is not an EV formula that I’m aware of.

Where did you find this formula?

This formula is depicting a return payout pattern of (0.97) on the first wager, and (0.97 * 0.97 = 0.9409) on the second wager, and (0.97 * 0.97 *.097 = 0.912673) on the third wager..., etc... etc..., thirty-six times in sequence.   This could happen but it’s very unlikely. It certainly isn’t an expectation for the above parameters.  A pay out like this would be a major disincentive to gamblers –exactly the opposite of what casinos want.   

GA

--. .-

... is my formula correct?

No. It’s not.

You are describing what appears to be a “Cashable Bonus”

The formula for this is...

EV = Bonus – (Wagering Requirement * House Edge)

Your example above (appears to be) ...

5000 bonus, 500,000 Wagering Requirement, 3% house edge. 

If so, then ...

\(EV = 5000 – (500000 * 0.03) = -10000\)

The expected value is a loss of 10,000 (dollars, pounds, drachmas). The loss is twice the bonus ... A fool’s gambit for sure.

\(EV = 0 \text{ at } 33 \dfrac{1}{3} \;{ turnovers}\)  . 

Note: This does not factor in bankruptcy (busting).  

Bankruptcy probability is based on the game probabilities, bank roll, and the minimum wager (bet). However, you still get the bonus if you bankrupt as long as it happens on or after the minimum number of turnovers.

GA

--. .-

The above solution and solution method is wrong.

...a box of berries having a total weight of 200 kg. The berries were 99% moisture, by weight. After two days in the sun, the moisture content of the berries was only 98%, by weight. What was the total weight of the berries after two days, in kg?

Solution:

\(\text {200 Kg of berries are 99% water. The weight of the water is then}\\ \left(0.99\cdot 200\right) = 198 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 200-0.98(200-x)\right)=x\\ \begin{aligned} 198-(196-0.98x)&=x\\ 198-196+0.98x&=x\\ 2+0.98x&=x\end{aligned}\\ \begin{aligned}1+0.98x-0.98x&=x-0.98x\\ 2&=0.02x\\ x&=100\\ \end{aligned}\\ \text { }\\ 200-x=200-100=100 ~ kg \leftarrow \text {The total weight of the berries after 2 days}\\ \)

GA

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May this might be because of Bing?

Why yes, Bing, the Microsoft search engine, causes all sorts of problems –on the net and everywhere else. It causes inflation, software crashes, constipation, diarrhea, PMS, and stupidity. It spreads viruses –not just computer viruses; it causes influenza and Covid outbreaks everywhere.  So, it most certainly causes problems with LaTex on Web2.0calc.com. 

I asked the Oracle of Bing how it is that you’ve managed to remain alive this long.  Fourteen hours later it presented the most probable reason:

DUMB LUCK.   

The Oracle of Bing did not define it as good luck or bad luck, because that would depend on the reader’s perspective. 

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On a serious note, I actually do understand your (implied) question, which is (maybe) how to edit or insert ascii text to the left of a rendered LaTex statement. The answer depends on which form of LaTex you are using.

For inline LaTex, finding the beginning of the LaTex is obvious. The Latex is unrendered, displayed as ASCII code while in edit mode; so, to the left of the delimiter that activates the LaTex is the spot. The rendered LaTex displays normally in preview mode. 

In your above post, you are using the LaTex dialog box (not to be confused with \mbox) for each LaTex statement. In this case, as you move your mouse over the LaTex statement string, the statement string will highlight with a yellow/amber outline. At this point, left click with your mouse and it will highlight with a light blue background and outline; then, using the curser key, move one space to the left. You can then type in what you want. If you double click the highlighted LaTex then it opens the dialogue box, where you can edit the LaTex.

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Some forum history...

Andre Massow updated the forum’s editor with the LaTex-finding outlines/highlights in past year or so. Prior to this, you had to search for a 10x1 pixel vertical line, which is the actual space occupied by the ASCII tag in the document. The only way to know you found it was to observe the word “math” appear on the bottom left side [body p math] while in edit mode. These boxes could sometimes be a royal pain-in-the-ass to find.  This is normally used to searched for and edit the LaTex in the dialogue box.

The above is not a long text. It may seem that way to you because you are still learning how to read and write Standard English.  You have a long way to go... Your communication skills are so poor and chaotic that is seems intentional. They are much more worser than your math skills. 

Understanding what you write often requires a large amount of effort. None of the AI assisted babble translators offer much help. I am marginally successful (or not) to a point because I have extensive, practiced skills in translating the dialects of moron, imbecile, and idiot. Even so, it is still a time consuming chore.   If you could consistently maintain at least a minimum moron-level of communication, it would be easier...   Not easy, just easier. 

Keep in mind that verbose is NOT the same as articulate.  Using made-up “English” words, whether you make them up yourself or they were taught to you, require several uses in various contexts before they can be translated consistently.  “Onliest” is one you’ve used in a post. If you use this a few more times, I should be able to translate this word in context.   

Here, you might be able to understand this more better:

It is not for me to demand English properly from writing by you. But I sugest now understanding without this is not completely possible without most all the possibilities examined.  So if you can give rise to more better possibilities, then my not demanding English properly from writing by you at the moment will require less demanding reporting when moments like this are in the future maybe. I’m not writing the new rules; I’m just explaining the rules that will help stand your writing more better comprehensible. Or, 

GA