Help Me Guess The Jellybeans!

[Infested Cats]

Originally posted by neevakee@ earlier
are the jelly beans jelly belly brand or are they a diffrent brand.  They would be marked on the bean.  If so I can say theat jelly bellys go for about $6 a pound.  If not I cant help you

I dont think the jellybeans are jellybelly. My school is too cheap to buy the "gormet" brand jellybeans...

Your guesses so far:
Got_memory? = 500
Darth Sushi = 1300
sara5564 = 800
dj7295 = 1280
Z = 700
Spindulik = 1140
Curley = 1450
merlin-1 = 1550

I'm pretty confident with your guesses! But, we should come to a final answer though!

[oldmancan]

IC this is probably too late to help. I suggest you go by the weights if you can. IF you can get an accurate container weight, big IF. The other weight you need is of one of the same brand jelly bean.

I think that production of the beans is pretty standardized, meaning all the beans will weigh the same. So if you can weigh the full container, an identical empty container and one jelly bean YOUR gonna win!! CHEATER!!

The thing about volume has been pointed out...settling. Good Luck.

[KinkoStinky]

1769

[Infested Cats]

Originally posted by oldmancan@16 April 2003 - 12:24
IC this is probably too late to help. I suggest you go by the weights if you can. IF you can get an accurate container weight, big IF. The other weight you need is of one of the same brand jelly bean.

I think that production of the beans is pretty standardized, meaning all the beans will weigh the same. So if you can weigh the full container, an identical empty container and one jelly bean YOUR gonna win!! CHEATER!!

The thing about volume has been pointed out...settling. Good Luck.

So, what you're saying is this:

Weigh the full container of jellybeans
Weigh an identical empty container
Weigh one jellybean

But what is the math from that point?

You subtract the weight from the container, then take what's remaining, and divide it by the weight of the one jellybean.

[KinkoStinky]

Not all jelly beans weigh the same...

[Infested Cats]

Ok, 2 of my friends and I weren't allowed to weigh the container. They said it would give us an unfair advantage (because no one else thought to do it).

Anyway, we tried a different method:

First, we counted how many jellybeans were along each diaganol-side. Then we counted how many there were from the top to the bottom, and from one side to the other. Then we took the surface area from the face and then we multiplied that by how many were across.

The math may not make sense, but i'll draw a picture later.

We came up with the number 1638. It seems pretty accurate considering all of your guesses. If we stick with that number, i'll go with 1638, my one friend will go 12 more, and the other will go 12 less.

What do you think? This is due TOMORROW! (4/16/03)

[KinkoStinky]

Your wronge...

[Z]

ok, take out mine and got memories guesses (700 and 500) and average the rest. then take your estimated guess and average it with the other average. then add or take away about 15 or 20, then presto!

[Infested Cats]

Originally posted by Infested Cats@16 April 2003 - 14:27
Ok, 2 of my friends and I weren't allowed to weigh the container. They said it would give us an unfair advantage (because no one else thought to do it).

Anyway, we tried a different method:

First, we counted how many jellybeans were along each diaganol-side. Then we counted how many there were from the top to the bottom, and from one side to the other. Then we took the surface area from the face and then we multiplied that by how many were across.

The math may not make sense, but i'll draw a picture later.

We came up with the number 1638. It seems pretty accurate considering all of your guesses. If we stick with that number, i'll go with 1638, my one friend will go 12 more, and the other will go 12 less.

What do you think? This is due TOMORROW! (4/16/03)

Me and my friends forgot to average in the amount of jellybeans that are between the lid and the rest of the container.

After finding the area of the area of the circle lid, we assumed about 50 more jellybeans were up in the "lid-area."

So, me and my friend's final number comes to about 1688. We're still going to do about 12 up (1700) and 12 down (1676).

[oldmancan]

IC were you suprised that you weren't allowed to weigh the container? lmao
Ya, you got the math, I knew you would. And your math seems reasonable so far as the latest method.
I know you haven't time but the best volume method would be to calculate the capacity (volume) of the container using Simpson's Rule. Something for you to search for. I think you want Simpsons 2nd rule. Have fun with that!!

King Yoshi> I suggest that the jelly beans from one production batch will be the same weight (on average) at least close enough for IC's purpose.