CPhill

y^2 - ay + 8by - 8ab = 0      factor by "grouping"

y(y - a) + 8b(y - a)  = 0    rewrite as....

(y - a)(y + 8b)  = 0    and setting each factor to 0, we have that......

y = a   or   y = -8b

Since 6.3, 6.5 and 6.9  are all repeated, we might consider this to be a "tri-modal" situation.......there a 3 "modes".....

However.......I think there is some debate about this........I believe some texts/instructors consider this data set to have no mode since no one value appears more than any of the others......

Maybe you should check to see which convention your text/instructor uses.......

The next multiple of 3 after 62 is 63..and 213 is the nearest multiple of 3 that is less than 215...the number of multiples is given by :

[(213 - 63) / 3 ] + 1  =    [150/3] + 1   = 50 + 1  = 51

Anonymous.......since we're multiplying, the order doesn't matter.......The usual convention is to write the factors in ascending order, but this isn't a hard and fast "rule."

Hey....that looks pretty good.....think you could sneak me a piece????

We can use trig to solve this....we have

tan(38) = x / 4.5      multiply both sides by 4.5

(4.5)tan(38) =  x =  about 3.5

Pi (square) is useful if you don't prefer Pi (round).......since......a square Pi is always easier to divide into equal portions than a round Pi .......

I guess we have to make one other assumption, here.....that a cannibal remaining in the boat actually isn't on the "side" of the river......if not....there's no way to solve this....to see why.....note that........

One the first crossing.......there are either....

a) Two cannibals in the boat and one gets let out......but now there is one more cannibal on a side of the river than a photog......and this violates the original condition

b) There are two photogs in the boat.....and this will clearly violate the original condition, since there are 3 cannibals and 1  photog at the starting point......

c) There is one cannibal and one photog in the boat.......the photog has to be let off......because, if the cannibal was let off, there again would be one more cannibal on a side than a photog

I actually think the problem means to say that, if there are cannibals and photogs together on either side of the river,  the cannibals cannot ever outnumber the photogs. Thus, we could have a situation where a lone cannibal (or two together) is/are on one side of the river and no photogs on that side ......(since there is nobody to eat in that situation).....!!!