Consider two different implementations of the same instruction set architecture. the instructions can be divided into four classes according to their cpi (class a, b, c, and d). p1 with a clock rate of 2.5 ghz and cpis of 1, 2, 3, and 3, and p2 with a clock rate of 3 ghz and cpis of 2, 2, 2, and 2. given a program with a dynamic instruction count of 1.0e6 instructions divided into classes as follows: 10% class a, 20% class b, 50% class c, and 20% class d, which implementation is faster? a. what is the global cpi for each implementation? b. find the clock cycles required in both cases.

just find the gcf and solve the gcf's until you get each answer i have a really good feeling somethings missing xd (like the dimensions of the fence) *cough *cough *hint *hint

step-by-step explanation:

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.