jelly50 jelly50
  • 03-04-2018
  • Mathematics
contestada

evaluate 10 sigma n=2 25(0.3)^(n+1)

Respuesta :

Sarith
Sarith Sarith
  • 20-02-2019

Answer: 0.964

Step-by-step explanation:

Create a series with the "n" values starting with n=2 until n=10: 25(0.3)^(2)+1 + 25(0.3)^(3)+1 + 25(0.3)^(4)+1 + 25(0.3)^(5)+1 + 25(0.3)^(6)+1 + 25(0.3)^(7)+1 + 25(0.3)^(8)+1 + 25(0.3)^(9)+1 + 25(0.3)^(10)+1

= 0.675 + 0.2025 + 0.06075 + 0.018225 + 0.0054675 + 0.00164025 + 0.00049207 + 0.00014762 + 0.00004428

≈ 0.964

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