Rom

\(\text{let }u = (a,c,e),~v = (b,d,f)\\ \text{we are given that }\\ u\cdot u + v\cdot v = 6\\ u\cdot v = 3\\ \text{and we are asked for max of }(u+v)\cdot (1,1,1)\)

\(\text{By Cauchy-Schwarz}\\ |(u+v)\cdot(1,1,1)|^2 \leq (u+v)\cdot (u+v) \times (1,1,1)\cdot (1,1,1) = \\ (u\cdot u + v\cdot v + 2u\cdot v)(3) = \\ (6+2(3))(3) = 36 \)

\(\text{so }(u+v)\cdot (1,1,1)\leq 6 \\ \text{with the maximum occurring at 6}\\ \text{thus the maximum of }a+b+c+d+e+f = 6\)

\(99x+100y+101z = 2009 \\ \text{mod both sides by 100}\\ z-x = 9 \\ z=x+9\\ 99x + 100y + 101(x+9) = 2009 \\ 200x + 100y + 909 = 2009\\ 200x + 100y = 1100\\ 2x+y=11\\ y=11-2x\)

\(\text{Now }y>0 \text{ so }1\leq x \leq 5\\ \text{thus we have the following 5 triplets}\\ (x,y,z) = (1,9,10),~(2,7,11),~(3,5,12),~(4,3,13),~(5,1,14)\)

\(P[\text{Joe eats at least two different kinds of fruit}] = \\ 1-P[\text{Joe only eats 1 kind of fruit}] = \\ 1 - 3\cdot \left(\dfrac 1 3\right)^3 = \\ 1 - \dfrac 1 9 = \\ \dfrac 8 9\)

You don't need to multiply all this out

Just look at the denominator.  It's not allowed to be zero.

\(d = (5k^2 - 25k) \cdot 7k^4 = \\ 35k^5(k-5)\\ \text{so }k\neq 0,~5\)

so (a) and (d)

1)

The orbit of the moon isn't perfectly regular

The Sun appears at different places with respect to Earth and Moon during different days and your model may not account for this.  (Though modern models do)

There may be local geography that manages to somehow pool or channel water in places afftecting the tidal height.

2)  It will be the fractional duration of the day that has passed.  It can be calculated (to the nearest second) as

\(\text{fractional part of the day} = \dfrac{\text{seconds since midnight}}{24\cdot 3600}\)

well you've ticked the right answer.. 

Is there a question?

a) no, these shouldn't depend on one another

b) yes, and we expect more depth with more rainfall, thus positive correlation

c) no, these shouldn't be correlated

d) these are correlated but we expect less of one with more of the other, i.e. a negative correlation.

The answer is (b)

boo CPhill, enough of these have been done that I'd have thought you'd seen one by now.

You need to take successive adjacent differences of f(x)

\(\left( \begin{array}{c} \{30,0,-2,0,30,160,510\} \\ \{30,2,-2,-30,-130,-350\} \\ \{28,4,28,100,220\} \\ \{24,-24,-72,-120\} \\ \{48,48,48\} \\ \end{array} \right)\)

in the matrix above the first line is f.

successive lines are formed by taking the line above minus it shifted 1 place to the left

that we could do this 4 times before all the numbers were equal shows that f is degree 4

A and B are independent events because P(A|B)=P(A) and P(B|A)=P(B) 

is correct

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