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Naddika [18.5K]
3 years ago
9

Point P is on line segment OQ. Given OQ = 4x + 2, PQ = 4x – 10, and

Mathematics
2 answers:
Verizon [17]3 years ago
7 0

Answer:

22

Step-by-step explanation:

Since, Point P is on line segment OQ.

\therefore O - P - Q\\\\\therefore OP + PQ = OQ\\\\\therefore 3x - 3 + 4x - 10 = 4x + 2\\\\\therefore 7x - 13 = 4x + 2\\\\\therefore 7x - 4x = 2 + 13\\\\\therefore 3x = 15\\\\\therefore x = \frac{15}{3}\\\\\huge \purple {\therefore x = 5} \\\\OQ = 4x + 2 = 4\times 5 + 2 = 20 + 2\\\\\huge \orange {\boxed {OQ = 22}}

Natali5045456 [20]3 years ago
7 0

Answer:

OQ=5

Step-by-step explanation:

The formula:

OP+PQ=OQ

Fill in the values for each segment using the values given:

3x-3+4x-10=4x+2

OP^   PQ^    OQ^

Simplify by adding like terms:

<u>3x</u>-3+<u>4x</u>-10=4x+2

7x<u>-3-10</u>=4x+2

add 13 to the right to isolate the variable:

7x<u>-13</u>=4x<u>+2</u>

   +13     +13

7x=4x+15

subtract 4x to the left to isolate the variable:

7x=<u>4x</u>+15

-4x -4x

3x=15

Divide both sides by 3 to get the variable by itself:

<u>3x=15</u>

3    3

x=5 so OQ=5

<u />

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6 0
3 years ago
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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