The first pump empties half the pond in 3 hours, so in 1/6 that time (1/2 hour), it empties (1/6)·(1/2) = 1/12 of the pond.
The second pump empties the other 5/12 of the pond in that half hour, so has a pumping rate of (1/2 h)/(5/12 pond) = (6/5 h)/pond.
The second pump could do the entire job alone in 1 hour and 12 minutes.
Let
x-------> total peanuts originally from the bag
we know that
1) Phillip took 1/3 of the peanuts from the bag--------> (1/3)*x
remaining=x-(1/3)*x-------> (2/3)*x
2) Joy took 1/4 of the remaining peanuts-------> (1/4)*[(2/3)*x]----> (1/6)*x
remaining= (2/3)*x-(1/6)*x------> (1/2)*x
3) Brett took 1/2 of the remaining peanuts------> (1/2)*(1/2)*x-----> (1/4)*x
remaining= (1/2)*x-(1/4)*x-------> (1/4)*x
4) Preston took 10 peanuts------> 10
(1/4)*x-10=71----> multiply by 4 both sides----> x-40=284----> x=324 peanuts
5) Total originally peanuts from the bag is equal to 324 peanuts
6) Phillip took (1/3)*x-----> (1/3)*324=108 peanuts
7) Joy took (1/6)*x------> (1/6)*324=54 peanuts
8) Brett took (1/4)*x------> (1/4)*324=81 peanuts
9) Preston took 10
so
check
108+54+81+10=253
remaining=324-253------> remaining=71-------> is correct
Step-by-step explanation:
so we're making two draws *with* replacement (this is important)
step 1: for the first draw, it wants the probability of getting a sour candy. to calculate this:
(# of sour candy) / (total # of candy)
step 2: for the second draw, it wants the probability of *not* getting a sour candy. to calculate this, you can calculate 1 - (the probability form part 1).
step 3: to find the probability of both events happening together, simply multiply the probabilities from part 1 and 2 together
side note: for step 2, you can only do this because the candy is being replaced. if there were no replacement, you'd have to re-calculate (# of non-sour candies) / (total after the first candy is drawn)
F(x) = (x - 4) (x^2 + 4) would be your answer.
