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svet-max [94.6K]
2 years ago
15

What is the upper and lower bound of 0.82

Mathematics
2 answers:
In-s [12.5K]2 years ago
8 0
Whatttt huh didnt ybderstand has
MrRa [10]2 years ago
6 0

Answer:

What ?

Step-by-step explanation:

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I believe  -7 + 91i. i hope this helps

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If it took 8 hours to clean 12 houses how many houses could be cleaned in 12 hours
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Q16. Find x^2 + 1/x^2 , if x + 1/x = 3
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Answer:

2 \times  x \times \frac{1}{x}  =  2 \\  {x}^{2}   +  {y }^{2}  + 2xy =  {(x + y)}^{2} \\ {x}^{2}  +  \frac{1}{ {x}^{2} }  + 2 = {(x +  \frac{1}{x} })^{2}  =  {3}^{2}  = 9

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Question Help Of 515515 samples of seafood purchased from various kinds of food stores in different regions of a country and gen
Setler79 [48]

Answer:

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

Step-by-step explanation:

Sample size, n = 51

p = 0.62

1 - p = 1 - 0.62 = 0.38

n = 515

Confidence level = 90% = Zcritical at 90% = 1.645

Confidence interval = (p ± margin of error)

Margin of Error = Zcritical * sqrt[(p(1-p))/n]

Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]

Margin of Error = 1.645 * 0.0214

Margin of Error = 0.035203

Lower boundary = (0.62 - 0.035203) = 0.584797

Upper boundary = (0.62 + 0.035203) = 0.655203

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

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