Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: CADENA DE MARKOV ESTABLE DE DIMENSIÓN 3. EJERCICIO DE APLICACIÓN EL PROBLEMA DE LOS SUPERMERCADOS En un pueblo hay tres supermercados { S1, S2, S3 } existe la movilidad de un cliente de uno a otro. Este mes, el 0.098 de los clientes va al S1, 0.39 al S2 y el resto al S3 Cada mes el S1 retiene el 0.66 de sus clientes y pierde el 0.25 que se va al S2 Se averiguó que el S2 solo retiene el 0.0026 y pierde el 0.91 que se va al S1 y el resto se va a s3 . El S3 pierde el 0.95 que va al S1 y el 0.019 que va al S2. Dada la matriz de transición M = [[0.66,0.25,0.094],[0.91,0.0026,0.088],[0.95,0.019,0.026]] y el vector de condiciones iniciales V = [0.098,0.39,0.51] , calcula: a) La potencia 2 de la matriz M es [M to the power of # n end exponent]={[[0.75,0.17,0.086],[0.68,0.23,0.088],[0.67,0.24,0.092]]; [[0.75,0.17,0.082],[0.69,0.23,0.084],[0.65,0.25,0.096]]; [[0.76,0.0064,0.23],[0.68,0.0016,0.32],[0.67,0.00054,0.33]]} b) El vector que representa las condiciones para ese estado 2 es V = {[0.18,0.2,0.21]; [0.18,0.2,0.21]; [0.2,0.23,0.23]} c) El vector de estabilidad es W = {{x=0.73,y=0.18,z=0.087}; {x=0.73,y=0.18,z=0.083}; {x=0.74,y=0.0049,z=0.26}}
Right answer summary: part 1: [[0.75,0.17,0.086],[0.68,0.23,0.088],[0.67,0.24,0.092]]; part 2: [0.18,0.2,0.21]; part 3: {x=0.73,y=0.18,z=0.087}
Question state: todo