Spaghetti in a bowl question. (pick up one end of a spaghetti and you can either join it to the other end of the spaghetti you are holding which created a loop or you can pick any other end in the bowl and join it to that. find the expected number of loops.) another spaghetti question (i think the interviewer loved spaghetti): you have a plate of spaghetti in front of you (no you pick two ends and tie them together. then you pick two more ends and tie them together. continue until there are no free ends left. if there were n spaghettis originally, what is the probability that you now have a single giant loop consisting of all the spaghettis?

i will with my problems and all the other things that need to   be on by the way

step-by-step explanation:

1/2(2*4)=6 $$- agahajakanajak

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.