Math Is Fun Forum

The discontinuity:

n,m,g,W,+
if we name the mid verticle cable the 0th verticle, we can also name all five verticle cables -2nd verticle, -1st verticle, 0th verticle, 1st verticle and 2nd verticle. and then we can define on each Δx is distributed with same mass m. g represents acceleration due to earth gravity.

1/2mg=def=W hence mg=2W

at the upper end of nth verticle(the tri-knit), forces are always 3, one downward(gravitional force), one leftward, and one rightward. define horizontally rightward and verticly upward +,and use cartesian expression of vectors. Using Newton's 1st and 2nd Law,we will easily get the table below.

nth                     0            1             2               3 (not displayed in graph)
leftwardforce (-h*,W)  (-h,-W)#  (-h,-3W) (-h,-5W)  *h is unkown
rightwardforce(h,W)#  (h,3W)     (h,5W)    (h,7W)
LF tangent         --          W/h         3W/h         5W/h
height-increament--     (W/h)Δx   3(W/h)Δx  5(W/h)Δx
height              0(set)  (W/h)Δx   4(W/h)Δx  9(W/h)Δx
distance           0(set)      Δx           2Δx           3Δx

# (h,W) and (-h,-W) are anti each other (Newton's 2nd Law)
at nth knit, LF+RF=2W, always(from 1st Law)
hence height= C distance², ignoring measure

but if the main cable behave more like a soft rope rather than a solid dome, it will bend more at the knits, thus not perfect parabola.

Thank you for your enthusiasm, jchristophm. It's really natural to think it as a catenary. (actually it was also my instinct)

Myorigenal model (edited):
My model is to cut the whole by the same horizontal distance, Δx,  and assume that cablepieces are straight, and that the concret road pieces are don't support each other. It finally derives the tangents  of the cable pieces develop by 0, k, 3k, 5k, 7k...to each side, where k is an unknown constant related to cable flexibilty. And that's similar to the motion that an arrow does without air resistance.

The solution from your link is more advanced.

Still, to be exact, allow me to say that it's only near parabola for in-continousity.

Any way, next time we see a SB, we will enjoy its beauty and know its cables rest almost parabolas.

2) 1/365 1/365[sup]2[/sup] is the probability that the 2 persons' birthday happen to be on a particular day, say April 1st, consider all 365 cases in particular days in a year satisfy the problem, the total probability would be
1/1/365[sup]2[/sup] (Jan 1st)+ 1/1/365[sup]2[/sup] (Jan 2nd) +...+1/1/365[sup]2[/sup] (Dec 31st)=1/365
this is one way to solve. an simplier alternative is to equate the second person's birthday to the given first one's 1/365

1)among the 15 drivers living on one particular street, 3 were involved in a car accident last year--ignore this piece of information.(for samples too few to prove the existing 8% probability wrong)
thus 8% is the chance that one single driver of the random 15 is involved in a CA
Let x denotes the number of drivers amoung 15 happened to be involved in a CA
P(0)+P(1)+ ... +P(15)=1
P(0)=0.92[sup]15[/sup]
P(1)=C(1,15) 0.08 0.92[sup]14[/sup]
P(2)=C(2,15) 0.08² 0.92[sup]13[/sup]
C(1,15)=15   C(2,15)=15 14 / 2!
C(k,m)=def= Number of Ways to Select k from m
the prob should be 1-P(0)-P(1)-P(2)

for q4 , my advice is --Check out the other 4 definations!

0.0 0.1 0.2 0.3 0.4  5 numbers
0.5 0.6 0.7 0.8 0.9  5 numbers
fair

0.00 0.01 ... 0.49 50 numbers
0.50 0.51 ... 0.99 50 numbers
fair

0.000 0.001 ... 0.499 500 numbers
0.500 0.501 ... 0.999 500 numbers
fair

so the rounding can be inferred (imagined) as fair.
when talking about infinity, you can only infer and define.

I like to graph...

Pythagoras' Theorem

Or did they really see some strange effect which they cannot duplicate because they don't know what caused it?
--Perhaps

Mind mapping??

Welcome lkomarci, you may post some of overchallenging questions here, and let us solve them together.We are not alone.

hope you won't be landing a rocket

you may compare its competence with common rounding and the half's exponents system 1/2 1/4 1/8 ...