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xenn [34]
3 years ago
12

#9. Solve: 8x + 5 = 3x + 15 X = 4 x = 2 x = 20/11 x = 20

Mathematics
1 answer:
Jobisdone [24]3 years ago
6 0

Answer:

x = 2

Step-by-step explanation:

8x + 5 = 3x + 15

8x - 3x + 5 = 3x - 3x + 15

5x + 5 = 15

5x + 5 - 5 = 15 - 5

5x = 10

5x / 5 = 10 / 5

x = 2

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A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de
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a= μ-3.16*σ , b= μ+3.16*σ if each fish caught is equally likely to be any one of these 4 distinct types.

<h3>What is meant by Chebyshev inequality?</h3>

Chebyshev's inequality is a probability theory that ensures that, over a vast range of probability distributions, no more than a particular proportion of values would be present within a selected limits or range as from mean. In other words, only a certain fish caught will be discovered within a given range of the distribution's mean.

The formula for which no more than a particular number of values can exceed is 1/K2; in other words, 1/K2 of a distribution's values can be more than or equal to K standard deviations away from the distribution's mean. Furthermore, it asserts that 1-(1/K2) of a distribution's values must be within, but not include, K standard deviations of the distribution's mean.

How to solve?

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y =  the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 →  1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90  →  1- σ²/(σ²+λ²)=  1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

In order to learn more about Chebyshev inequality, visit:

brainly.com/question/24971067

#SPJ4

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