Alice and bob play the following game: they start with an empty 2008x2008 matrix (p. s. take a wild guess which year this was) and take turns writing numbers in each of the 20082 positions. once the matrix is filled, alice wins if the determinant is nonzero and bob wins if the determinant is zero. if alice goes first, does either player have a winning strategy?

Bob wins

Step-by-step explanation:

Think first in a simpler case. In a 2x2 matrix the determinant is zero if two columns or rows are proportional. So Alice writes the first number, then Bob write the same number in the same column but in the other row. Bob repeat the procedure after Alice's second number, and win.

If the matrix is nxn where n is even (like 2008) Bob will win following the previous procedure. Alice writes a number anywhere, Bob writes the same number in the same column in the above (or below) row.  Every time Alice write a number in the same row as her first one, Bob copy that number in the above (or below) row. Any other rows are not important, i.e, the determinant will be zero with any values in them.    

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.