Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: CADENA DE MARKOV ESTABLE DE DIMENSIÓN 3. EJERCICIO DE APLICACIÓN Dada la matriz de transición M = [[0.78,0.031,0.19],[0.61,0.069,0.32],[0.24,0.082,0.68]] y el vector de condiciones iniciales V = [0.97,0.028,0.0061] , calcula: a) La potencia 5 de la matriz M es [M to the power of # n end exponent]={[[0.57,0.051,0.38],[0.56,0.052,0.38],[0.54,0.055,0.41]]; [[0.8,0.041,0.15],[0.8,0.04,0.16],[0.81,0.041,0.15]]; [[0.56,0.035,0.41],[0.55,0.036,0.41],[0.53,0.039,0.43]]} b) El vector que representa las condiciones para ese estado 5 es U = {[0.56,0.55,0.52]; [0.78,0.78,0.78]; [0.54,0.54,0.52]} c) El vector de estabilidad es W = {{x=0.55,y=0.037,z=0.42}; {x=0.56,y=0.053,z=0.39}; {x=0.81,y=0.041,z=0.15}}
Right answer summary: part 1: [[0.57,0.051,0.38],[0.56,0.052,0.38],[0.54,0.055,0.41]]; part 2: [0.56,0.55,0.52]; part 3: {x=0.56,y=0.053,z=0.39}
Question state: todo