Suppose you have some money to invest-for simplicity, $1-and are planning to put a fraction w into a stock market mutual fund and the rest,1-w , into a bond mutual fund. suppose that a $1 invested in a stock fund yields rs after one year and a $1 invested in a bond fund yields rb. rs and rb are random variables with expected value of 8% and 7% respectively, and standard deviation of 7% and 4% respectively. the correlation between rs and rb is 0.25. if you place a fraction w of your money in the stock fund and the rest,1-w, in the bond fund then the return on your investment will be r = wrs + (1-w)rb . a. compute the mean and standard deviation of r for w = .5 and w = .75 respectively. b. what value of w makes the mean (or expected value) of r as large as possible? what is the standard deviation of r for this value of w? c. what is the value of w that minimizes the standard deviation of r? (you can show this using algebra, or calculus).

Asystem of linear equations with more equations than unknowns is sometimes called an overdetermined system. can such a system be consistent? illustrate your answer with a specific system of three equations in two unknowns. choose the correct answer below. a. yes, overdetermined systems can be consistent. for example, the system of equations below is consistent because it has the solution nothing. (type an ordered pair.) x 1 equals 2 comma x 2 equals 4 comma x 1 plus x 2 equals 6 b. no, overdetermined systems cannot be consistent because there are fewer free variables than equations. for example, the system of equations below has no solution. x 1 equals 2 comma x 2 equals 4 comma x 1 plus x 2 equals 12 c. yes, overdetermined systems can be consistent. for example, the system of equations below is consistent because it has the solution nothing. (type an ordered pair.) x 1 equals 2 comma x 2 equals 4 comma x 1 plus x 2 equals 8 d. no, overdetermined systems cannot be consistent because there are no free variables. for example, the system of equations below has no solution. x 1 equals 2 comma x 2 equals 4 comma x 1 plus x 2 equals 24