Maths- Gym for Brainiacs
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Re: Maths- Gym for Brainiacs
length of what?Immortal Babun wrote:You don't compare perimeters you count the length
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Re: Maths- Gym for Brainiacs
Length of all the white squares you get.spanky489 wrote:length of what?Immortal Babun wrote:You don't compare perimeters you count the length
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Re: Maths- Gym for Brainiacs
zomg thank god this thread exists, i pretty please need your help
ok so ive got an assignment for stats and pretty much youve got to write a report for an imaginary business and they give you some data to base it from
anyway the imaginary company is a hotel and im trying to figure out the average amount each person spends in excess of their accomodaion fees per day, thing is im not sure which method to use
(the data gives you info for 200 bookings and in each booking amoung other things it tells you how many days each booking stayed for and their TOTAL spend above the accomodation costs)
anyway the first method i use was to divide the total spend of each booking by the durration of their booking and then find the average of the number which equals 235.8114167
the second method i used was to find the average total spend above accomodation costs (1147.775) and the average number of days spent at the hotel (4.785) and then go 1147.775/4.785 which comes to 239.8694
i get the feeling the first method is right but i really dont know, any help would be much appreciated, thank you
ok so ive got an assignment for stats and pretty much youve got to write a report for an imaginary business and they give you some data to base it from
anyway the imaginary company is a hotel and im trying to figure out the average amount each person spends in excess of their accomodaion fees per day, thing is im not sure which method to use
(the data gives you info for 200 bookings and in each booking amoung other things it tells you how many days each booking stayed for and their TOTAL spend above the accomodation costs)
anyway the first method i use was to divide the total spend of each booking by the durration of their booking and then find the average of the number which equals 235.8114167
the second method i used was to find the average total spend above accomodation costs (1147.775) and the average number of days spent at the hotel (4.785) and then go 1147.775/4.785 which comes to 239.8694
i get the feeling the first method is right but i really dont know, any help would be much appreciated, thank you
fatman123- Fan Favorite
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Re: Maths- Gym for Brainiacs
I think the first one because you want the average of the people, not the group?
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Re: Maths- Gym for Brainiacs
Is it your exact assignment? The question is very important because different statistics give different outlooks on a situation. You should know for sure what is asked from you. So what you calculated ( different outlooks) :fatman123 wrote:zomg thank god this thread exists, i pretty please need your help
ok so ive got an assignment for stats and pretty much youve got to write a report for an imaginary business and they give you some data to base it from
anyway the imaginary company is a hotel and im trying to figure out the average amount each person spends in excess of their accomodaion fees per day, thing is im not sure which method to use
(the data gives you info for 200 bookings and in each booking amoung other things it tells you how many days each booking stayed for and their TOTAL spend above the accomodation costs)
anyway the first method i use was to divide the total spend of each booking by the durration of their booking and then find the average of the number which equals 235.8114167
the second method i used was to find the average total spend above accomodation costs (1147.775) and the average number of days spent at the hotel (4.785) and then go 1147.775/4.785 which comes to 239.8694
i get the feeling the first method is right but i really dont know, any help would be much appreciated, thank you
1. The total average of how much a person spends in excess.
2. You go more into details, you specify how much the average excess is related to average stay of customers.
In second version, you got a bigger number which implies that some individuals stay significantly longer than others while some of the short stayers spend way more above their accomodation fees than others.
For example, you've got a Sheikh who stays for two nights only but spends shitloads of money and a businessman who stays for 2- weeks but spends much less.
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Re: Maths- Gym for Brainiacs
Waitwhat, that wasn't what I thought he meant...
Fatman, can you explain your methods clearer?
Fatman, can you explain your methods clearer?
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Re: Maths- Gym for Brainiacs
Immortal Babun wrote:Is it your exact assignment? The question is very important because different statistics give different outlooks on a situation. You should know for sure what is asked from you. So what you calculated ( different outlooks) :fatman123 wrote:zomg thank god this thread exists, i pretty please need your help
ok so ive got an assignment for stats and pretty much youve got to write a report for an imaginary business and they give you some data to base it from
anyway the imaginary company is a hotel and im trying to figure out the average amount each person spends in excess of their accomodaion fees per day, thing is im not sure which method to use
(the data gives you info for 200 bookings and in each booking amoung other things it tells you how many days each booking stayed for and their TOTAL spend above the accomodation costs)
anyway the first method i use was to divide the total spend of each booking by the durration of their booking and then find the average of the number which equals 235.8114167
the second method i used was to find the average total spend above accomodation costs (1147.775) and the average number of days spent at the hotel (4.785) and then go 1147.775/4.785 which comes to 239.8694
i get the feeling the first method is right but i really dont know, any help would be much appreciated, thank you
1. The total average of how much a person spends in excess.
2. You go more into details, you specify how much the average excess is related to average stay of customers.
In second version, you got a bigger number which implies that some individuals stay significantly longer than others while some of the short stayers spend way more above their accomodation fees than others.
For example, you've got a Sheikh who stays for two nights only but spends shitloads of money and a businessman who stays for 2- weeks but spends much less.
what bugs me most about this assignment is that there is no one question that i need to answer, its completly opened ended and you need to tackle it in your own manner, having said that the assignmnt does tell you that one of the businesses key performance indicators was that "the average customer spends more then $260 per day in excess of accommodation fees" now i know they havent reached that its just im not sure what the average spend/day in excess of accomodation fees actually is
ok to explain my methods more clearly:
1) TOTAL amount spent in excess of accomodation expenses/number of nights spent at the resort (do that for all 200 bookings) and then find the average of all those numbers which comes to 235.8114167
2) find the average total spend in excess of accomodation costs (ie sum of all total spend in excess of accomodation costs/200) and then find the average number of nights spend at the resort (sum total number of nights/200) which gives 1147.775 and 4.785 respectivley and then divide 1147.775 by 4.785 to give 239.8694
hopefuly that helps and thanks again to everyone
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Re: Maths- Gym for Brainiacs
The first version then. In which class are you?
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Re: Maths- Gym for Brainiacs
cheers babunImmortal Babun wrote:The first version then. In which class are you?
im studying economics at uni, but this assigment is for "business and economic statistics"
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Re: Maths- Gym for Brainiacs
Nice thread
I Maths
I Maths
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Re: Maths- Gym for Brainiacs
can you guys help me out?Im having some trouble completing this equation
3^2x-6.3x+9=0
i know i have to transform 3x in y,but i cant figure out how to get the 3x out of 3^2x
3^2x-6.3x+9=0
i know i have to transform 3x in y,but i cant figure out how to get the 3x out of 3^2x
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Re: Maths- Gym for Brainiacs
Does 6.3 mean 6 X 3?
And do you have to solve for x?
And do you have to solve for x?
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Re: Maths- Gym for Brainiacs
yes its 6x3,we just do "." here in brazil to not get it confused with the X.
and yeah,i need x
and yeah,i need x
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Re: Maths- Gym for Brainiacs
Is it 3^(2.x) or (3^2).x or (3.x)^2?
Last edited by kiranr on Thu Apr 12, 2012 3:25 am; edited 1 time in total
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Re: Maths- Gym for Brainiacs
Use Wolfram|Alpha. A must for students of mathematics.
http://www.wolframalpha.com/
Put "Solve," before the equation, and click on "Show steps," in the Result panel and it will briefly show the steps, though I don't think it's in much detail.
Oh, and use * for multiply.
http://www.wolframalpha.com/
Put "Solve," before the equation, and click on "Show steps," in the Result panel and it will briefly show the steps, though I don't think it's in much detail.
Oh, and use * for multiply.
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Re: Maths- Gym for Brainiacs
Ganso wrote:can you guys help me out?Im having some trouble completing this equation
3^2x-6.3x+9=0
i know i have to transform 3x in y,but i cant figure out how to get the 3x out of 3^2x
Let 3^x = y
therefore you get y^2-3y+9=0
then solve using whatever you like (quad formula i think)
then simplify to its easiest form, then replace the 'y' with 3^x and you get your answer. Should be two answers.
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Re: Maths- Gym for Brainiacs
Is it 3^2x -6*3^x +9 = 0 or 3^2x - 6*3x +9 =0 ?Ganso wrote:can you guys help me out?Im having some trouble completing this equation
3^2x-6.3x+9=0
i know i have to transform 3x in y,but i cant figure out how to get the 3x out of 3^2x
For the first version, use what ausbaz said, you'll get x1= sqrt(3) and x2= -sqrt(3).
For the second version, I changed the equation to:
3^2x = 18x-9 / :9
3^(2x-2)=2x-1
I drew the graphs with my prog, the interception points are the ones you're looking for. The interception takes place at x=1.
Last edited by Immortal Babun on Fri Apr 13, 2012 5:48 am; edited 1 time in total
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Re: Maths- Gym for Brainiacs
its the second one
i canot use programs in my test lol and i have to somehow change 3^x into y.
@ausbaz:i have no idea what you did there mate lol i think you didnt get what i wrote
I'll just ask my teacher next week,thanks though
i canot use programs in my test lol and i have to somehow change 3^x into y.
@ausbaz:i have no idea what you did there mate lol i think you didnt get what i wrote
I'll just ask my teacher next week,thanks though
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Re: Maths- Gym for Brainiacs
If you change 3^x into y, then it goes like this,
3^x = y
x * ln(3) = ln(y)
x = ln(y)/ln(3)
Then the equations becomes
y^2 - 17ln(y) + 9 = 0
Then i think you have to use some numerical approximation techniques to solve this.
3^x = y
x * ln(3) = ln(y)
x = ln(y)/ln(3)
Then the equations becomes
y^2 - 17ln(y) + 9 = 0
Then i think you have to use some numerical approximation techniques to solve this.
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Re: Maths- Gym for Brainiacs
what kind of equation is that lol
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Re: Maths- Gym for Brainiacs
i have no idea what ln is
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Re: Maths- Gym for Brainiacs
natural log
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Re: Maths- Gym for Brainiacs
Ganso wrote:i have no idea what ln is
i guess the only way to solve this problem analytically is to use a natural logarithms or (logarithms with the base of 3) and considering u dont know what they are i think you should double check what ur looking for and how.
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Re: Maths- Gym for Brainiacs
i dont even get the equation, nothing i've seen in my life and i have seen A LOT of equations
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Re: Maths- Gym for Brainiacs
i know what logarithms are i just use log and not ln,and thats not how you're supposed to do it,as i said,you are supposed to turn that into a ax^2+bx+c=0 quadratic equation and get the solution(s)
btw i just checked the answer and babun's is right,its 1,but i have to figure out a way of doing it on paper.im sure its not using logs because my teacher didnt even review logs yet
btw i just checked the answer and babun's is right,its 1,but i have to figure out a way of doing it on paper.im sure its not using logs because my teacher didnt even review logs yet
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Re: Maths- Gym for Brainiacs
RealGunner wrote:i dont even get the equation, nothing i've seen in my life and i have seen A LOT of equations
You see such equations as f(x) = x + ln(x) and these can be found while doing differentiation, integration, finding maxima, minima etc...
@Ganso, there is no way you can make that equation into a* (y^2) + b*y + c = 0.
Because, in your equation, the first term is 3^(2.x) and the second term is 18.x
So you can see that the first term has x in the exponent and the second term has x in the base. You have to use log or natural log.
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