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Fiesta28 [93]
3 years ago
15

Joseph's parents have planted two gardens. One is square and has an area of 25 ft 2. The other one has two sides equal to 2/3 of

one side of the square, and the other two sides equal to 5/2 of one side of the square.
A. Find the dimensions of the other garden. Explain how you found your answer.
B. Find the area of the other garden.
Mathematics
1 answer:
Tomtit [17]3 years ago
6 0
A. 15ft - 12ft

b.120ft2
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Find the sum. 2 3/<br>6 + 4 1/6​
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Answer:

6 4/6 =6 2/3

Step-by-step explanation:

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1990 population was 254 million in 2014 it was 303 what was the percentage increase in the population
Ann [662]

Answer:

19.29%

Step-by-step explanation:

Given

Population in 1990

= 254000000

Population in 2014

= 303000000

Increase = 303000000 - 254000000

= 49000000

% increase in population =

Increase /initial population x 100%

That’s

49000000/254000000 x 100%

0.1929 x 100%

19.29%

The population increased by 19.29%

7 0
2 years ago
What is p ÷ 6 = 9 ? I'm not really understanding it
Shalnov [3]
Let's figure this out. Step by step:

P /  6 = 9

P = 6(9)

P = 54 is your answer.

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8 0
3 years ago
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What is each product? For 5/10•10/3
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10 cancels out so its 5/3 :) Rate Me plz <span />
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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
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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