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pickupchik [31]

3 years ago

13

Given frequent itemset l and subset s of l, prove that the confidence of the rule "s’ => (l - s’)" cannot be more than the co

nfidence of "s => (l - s)", where s’ is a subset of:_________

1 answer:

aniked [119]3 years ago

4 0

Answer:

the answer is I

Step-by-step explanation:

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Please help I am so lost!!!

ASHA 777 [7]

$\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\
-----------------------------\\\\
tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\

\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\

\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}
\end{cases}\\\\$

$\bf -----------------------------\\\\
\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}
\\\\\\
\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies 
\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)$

4 0

3 years ago

Which scatterplot shows the weakest negative linear correlation?

kherson [118]

Answer:

the first option, i took the test

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Select the correct answer. Which graph represents the solution to the inequality?

melisa1 [442]

Answer:

A

Step-by-step explanation:

First you simplify both sides to get -2.4x+14.4 > 52.8

then you need to subtract the 14.4 from both sides giving -2.4x> 38.4

then divide both sides by -2.4 giving an answer of

x is less than or equal to -16 meaning you point to the left with a closed circle

8 0

3 years ago

solve the system of linear equations by elimination, the equations are 2x-6=4y and 7y= -3x+9. What is the solution

tatiyna

This should be the answer. You can plug in 0 for y to any of the equation to find the x.

3 0

3 years ago

The following gives information about the proportion of a sample that agree with a certain statement. Use StatKey or other techn

marishachu [46]

Answer:

(0.2278, 0.3322)

Step-by-step explanation:

Given that out of 400 people 112 agree and 288 disagree

Proportion of people agreeing = $\frac{112}{400} =0.28$

$q=1-p =0.72\\se = \sqrt{\frac{pq}{n} } =0.03175$

For confidence interval 90% we have critical value as

1.645

Margin of error =1.645*SE

= $1.645(0.03175)\\= 0.0522$

Confidence interval $= (0.28-0.0522, 0.28+0.0522)\\= (0.2278, 0.3322)$

3 0

2 years ago