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adell [148]
3 years ago
9

I need help with 40 and 39!!!

Mathematics
2 answers:
Inessa [10]3 years ago
4 0

Answer:

39) Opt. A (24)

Step-by-step explanation:

39) 40 - 8(n) where n = 2

=> 40 - 8*2

=> 40 - 16

=> 24 Opt. A

And just for information, the answer you marked for 38th is wrong.

Formula = bh/2

b = 16

h = 12

bh/2 = 16*12/2

        =192/2 = 96

Hope this helps u

Please mark as brainliest

Thank You

vagabundo [1.1K]3 years ago
3 0
39-A
When you multiply 8 times 2 its 16 and 40-16 is 24
40-D
less than is towards the negatives and for less than or greater than you use open circles
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Explain how you know 21/100 is greater than 1/5
Nataly_w [17]

This problem can be solved using equivalent fractions. The first step in resolving this problem is to realize that fractions are best compared when they both have the same denominator.  In this case, I will choose to make that common denominator 100. There is no need to rewrite the fraction \frac{21}{100} as the denominator for this fraction is already 100. The fraction \frac{1}{5} =\frac{20}{100} . This is achieved by multiplying both the numerator and denominator by 20. Now that the two fractions have the same denominator, we can easily see that \frac{21}{100} is greater than \frac{1}{5} because it is greater than its equivalent fraction.

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First answer gets a thanks
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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
gregori [183]

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

4 0
1 year ago
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