Let the process of getting through undergraduate school be a homogeneous Markov process with time unit one year. The states are Freshman, Sophomore, Junior, Senior, Graduated, and Dropout. Your class (Freshman through Graduated) can only stay the same or increase by one step, but you can drop out at any time before graduation. You cannot drop back in. The probability of a Freshman being promoted in a given year is .8; of a Sophomore, .85; of a Junior, .9, and of a Senior graduating is .95. The probability of a freshman dropping out is .10; of a Sophomore, .07; of a Junior, .04; of a Senior, .02. a) Construct the Markov transition matrix for this process.
b) If we were more realistic, and allowed for students dropping back in, this would no longer be a Markov process. Why not?

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step-by-step explanation:

add 6% to the value she got from selling the house.

-2 is the right

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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