Reminds me of tv ads:
“past performance is not indicative of future earnings.”
you guys are right, the 2 second lap has no time left to be completed.
I you want to go 1 mile at 60 mph then you have to do the whole mile in .0166 hours.
Well the first lap, .5 miles at 30mph took you .0166 hours, leaving no time left.
Unless, of course your talking low quantities of 1. (joke)
We live in an age that reads to much to be wise, and thinks too much to be beautiful–Oscar Wilde
Zyada, the user doesn’t know what he’s talking about. The averages will be the same. Here’s the proof:
Assume a series of daily amounts, call the total of these amounts “T”. Call the number of days “N”. Call the daily average “X”.
X = T/N
Now assume you want to project a new daily average for an additonal number of days “M” and you use the previously calculated value “X” as the assumed daily average for these days. We’ll call this second daily average “Y”.
Y = (T + MX)/(N + M)
The question is Y equal to X?
Y = (T + MX)/(N + M) {given above}
Y = (T + M(T/N))/(N + M) {substituting T/N for X}
Y = (T + MT/N)/(N + M) {multiplying the factors}
Y = (NT + MT)/N(N + M) {multiplying everything by N}
Y = T(N + M)/N(N + M) {seperating out T}
Y = T/N {dividing everything by (N + M)}
Therefore Y = X {both are equal to T/N}
Let x[sub]n[/sub] = datum from day n
A = average of first n days
then
- A = (x[sub]1[/sub]+x[sub]2[/sub]+ … + x[sub]n[/sub])/n*
Multiplying by n:
nA = x[sub]1[/sub]+x[sub]2[/sub]+ … + x[sub]n[/sub]
Now, your user wants you to assume x[sub]n+1[/sub], x[sub]n+2[/sub], etc. are the same as A. Adding A to the left side and x[sub]n+1[/sub] to the right gives:
nA + A = x[sub]1[/sub]+x[sub]2[/sub]+ … + x[sub]n[/sub]+x[sub]n+1[/sub]
Factoring the left side:
(n+1)A = x[sub]1[/sub]+x[sub]2[/sub]+ … + x[sub]n[/sub]+x[sub]n+1[/sub]
Dividing by n+1:
A = (x[sub]1[/sub]+x[sub]2[/sub]+ … + x[sub]n[/sub]+x[sub]n+1[/sub])/(n+1)
The right side is the formula for the average of the first n+1 data. The left is still A, the average of the first n. But it’s also the average of the first n+1. And it will remain the average no matter how many new data are added in, as long as they’re assumed to be equal to A.
“… mathematical finagling that would constipate Einstein…”
Ok maybe I am just not getting it here…
If you run a lap of a track (1 mile) at 30mph and you have to run 2 laps and have and average lap speed of 60mph where does the time factor come in?? or did I miss something?
The way I see it it breaks doen like this:
(30mph/2)+( X/2) =60mph right?
The average of 2 laps. So multiply both sides of the problem by 2 and you have:
30mph + x = 120mph solve for X
There are only two things that are infinite…the Universe and Man’s stupidity…I’m not sure about the Universe though.
You are in a two-lap race around a one mile track. The first lap you average 30 mph. How fast do you have to go on the second lap to average 60 mph overall?
*Patricinus Scriblerus: Ok maybe I am just not getting it here…
If you run a lap of a track (1 mile) at 30mph and you have to run 2 laps and have and average lap speed of 60mph where does the time factor come in?? or did I miss something?
The way I see it it breaks doen like this:
(30mph/2)+( X/2) =60mph right?
The average of 2 laps. So multiply both sides of the problem by 2 and you have:
30mph + x = 120mph solve for X*
First, figure how long it will take to do one lap at 30 mph:
<code>Formula: Rate * time = distance(r[sub]1[/sub]*t[sub]1[/sub]=d[sub]1[/sub]) </code>
(30 miles/hour) t[sub]1[/sub]=1 mile*
Dividing by (30 mph) gives:
t[sub]1[/sub] = 1 mile/30 mph = 1/30 hr = 2 minutes
Now, figure how long it will take to do two laps at an average speed of 60 mph:
*r[sub]2[/sub]*t[sub]2[/sub]=d[sub]2[/sub]
(60 mph)t[sub]2[/sub] = 2 miles
Dividing by (60 mph):
t[sub]2[/sub] = 2 miles/60 mph = 1/30 hour = 2 minutes
Now, the difference between t[sub]1[/sub] and t[sub]2[/sub] is the time you need to complete the second lap to make the whole trip’s average speed 60 mph.
t[sub]2[/sub]-t[sub]1[/sub] = 2 minutes - 2 minutes = 0 minutes.
Therefore, you have no time left to complete the second lap to make an average trip speed of 60 mph.
zyada –
Sorry for hijacking your thread. I thought my puzzler was an old one that everyone knew, or at least it was pretty obvious once it was explained (It’s taking longer than we thought).
But it looks like you got the answer you needed. Adding additional average data points to a computed average doesn’t change the average.
“If ignorance were corn flakes, you’d be General Mills.”
Cecil Adams
The Straight Dope
Pat Scribble:
You set the problem up this way:
(30mph/2)+( X/2) =60mph right?
Right? No, wrong. Simply first note that you should be equating numbers of time units. If one lap = n miles, your time equation should be:
(n[/n] mi / 30 mi/hr) + (n mi / x mi/hr) = 2n* / 60 mi/hr
The 'n’s cancel out, of course. Then you have:
1/30 + 1/x = 2/60
Solving that for x,
x = <font size=3 face=“symbol”>¥</font>
Ray (Is it true that he who laps last cleans up?)
Well, that first equation should’ve been:
(n mi / 30 mi/hr) + (n mi / x mi/hr) = 2n / 60 mi/hr
Ray
I’m with pluto, y’all are making this harder than it should be. It’s like, Car1 goes around the track twice at 60mph it takes 2 minutes, Car2 goes once around the track at 30mph it takes 2 minutes. Now Car2, floor it and match Car1’s time, oops, times up, sorry.
These are the errors:
What is the first number? Miles? Then you’re multiplying miles times miles per hour. That makes square-miles per hour. Bzzzzt!.
Since you’ve got mph/2 on one side of the equation and mph on the other, I suppose the 2 is an absolute number. So where did it come from? Where in the original problem does it say “divided by two”? Bzzzt!
The right setup is:
2miles/(1mile/30mph+1mile/xmph)=60mph
Checking the units:
miles/(miles/mph+miles/mph)=mph
miles/(miles/(miles/hours)+miles/(miles/hours))=miles/hours
miles/(miles(hours/miles)+miles(hours/miles))=miles/hours
miles/(hours+hours)=miles/hours
miles/hours=miles/hours
Good. Proceeding to the math:
2 / (1/30 + 1/x) = 60
1 / (1/30 + 1/x) = 30
(1/30 + 1/x) = 1/30
1/x = 0
x = 1/0
Ding!
Moral – always check your units!
John W. Kennedy
“Compact is becoming contract; man only earns and pays.”
– Charles Williams
::leafing through the school directory checking for math tutors::
Getting back to averages, joke in today’s comics section: Three statisticians go duck hunting. First statistician misses a duck too high, the second misses the duck too low. The third yells out, “It’s a hit!”
As to the problem, getting the units right is what got me out of Physics in college <lol>.