Why are there so many extremely basic math questions on Quora?
Because they’re trivial to write, and there’s enough folk here who like maths that they’ll get high viewer counts.Which means more $$$ if you’re trying to monetize paid-for questions. This is almost certainly the case for most such questions, as when you look at the questioner they’ve often asked hundreds of these with only minor variations.It’s just gaming the system in response to a perverse incentive.Worse still, you can’t easily block them because that then risks fewer similarly-tagged genuinely interesting maths questions in your feed. I do make a point of always reporting such questions as either spam or insincere though, and citing abuse of the partner program. If enough people do this, the cost of the manual effort to review all those reports might finally be enough to persuade them to try and fix the problem properly.Why, oh why, oh why, did they not make the partner programme all about the best answers instead? Quora would have improved so much as a result!
Why is it that I understand extremely complex math problems but I can't understand simple ones?
First of all I think you have a brain whose challenge mode is always On :PA simple question creates disinterest within you and hence you don't tend to solve for sake of anything. I guess you always set a higher standard for yourself in brainy activities. So necessarily you feel more competitive upon seeing a complex question and try to give your heart and soul to solve it. And if you are capable enough you solve that complex ques with full sincerity. Vividly you have great imagination power,deep thoughts, out of box thinking.
Can anyone help me with my math lesson? I'm extremely confused on what I'm supposed to do.?
1) I don't get it sorry. 2) If the original price is $8 and he gets $2 off then it's $6. 3)47.70 divided by 12 is 3.975 round that up to $3.98 each magazine. 4)47.70-26.40=$21.40 5) 26.40/12 is $2.20 so 3.98-2.20 = $1.78 6)1.15 x 6 = $6.90 7) Assuming there's 30 days in a month = 1.15 x 30 = $34.50 8) 2.25 + 0.75 = $3.00 9)2.25 x 3 = 6.75 + 2.25 = $9.00 10) $9.00 x 30 = $270 11) 8.50 x 2 = $17 12) 5.00+1.50+2.00 = $8.50 13) 17 + 8.50 = $25.50 14) 25.50 x 4 = $102 15) 204.6 - (all these answers added together) 16) Can he? This was really simple math.
What are some examples of deceptively simple-looking math problems?
In a slightly different light, I'm going to present a problem that seems simple but is undecidable in general. The general question is simple: given an arbitrary polynomial with integer variables, does it have a solution?Sometimes, it's quite easy:[math] x^3 + y^3 + z^3 = 29 [/math]You can probably do it in your head: [math] \begin{align} x &= 1 \\ y &= 1 \\z &= 3 \\ \end{align} [/math]Sometimes it's more tricky:[math] x^3 + y^3 + z^3 = 30 [/math]Can you figure out a solution to this one? You can't! But other people did, although it took until 1999!Why? This is why:[math] \begin{align} x &= -283059965 \\ y &= -2218888517 \\ z &= 2220422932 \\ \end{align} [/math]Yeah.So what about this one?[math] x^3 + y^3 + z^3 = 33 [/math]That's still an open problem.In fact, it turns out there's no algorithm to solve these kinds of equations. Just like the (in)famous halting problem, this is undecidable. The problems look so simple, but they're not just unsolved—they're fundamentally unsolvable! How's that for difficult?I originally read about this in some slides I can't find anymore. Happily, there's a great article about the topic from a Berkeley professor (now at MIT): "Undecidability in Number Theory".
Any Math help please?
In this exercise, use the following information: A supermarket polled 1,000 customers regarding the size of their bill. The results are given in the figure below. Use probability rules (when appropriate) to find the relative frequency with which a customer's bill is as stated. Size of Bill Number of Customers below $20.00 211 $20.00-$39.99 111 $40.00-$59.99 183 $60.00-$79.99 176 $80.00-$99.99 197 $100.00 or above 122 _____________ Do the following. (Round the answers to six decimal places.) (a) Find the probability of being dealt a "kings over queens" full house (three kings and two queens). (b) Find the probability of being dealt a full house. please, and thank you(: