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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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Determine the slope of the line that contains the given points <br><br> J(-5, -2), K(5, −4)

Irina-Kira [14]

Answer:

$-\frac15$

Step-by-step explanation:

Hello!

We can utilize the slope formula to find the slope.

Slope Formula: $S = \frac{y_2-y_1}{x_2-x_1}$

Remember that a coordinate is written in the form (x,y)

<h3>Find the Slope</h3>

$S = \frac{y_2-y_1}{x_2-x_1}$
$S = \frac{-4-(-2)}{5-(-5)}$
$S = \frac{-2}{10}$
$S = -\frac15$

The slope of the line is $-\frac15$ .

4 0

2 years ago

Read 2 more answers

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

3 years ago

Will mark brainliest for any help!

elena55 [62]

Answer:

5000

Step-by-step explanation:

6 0

3 years ago

1. What does it mean that a <br> reflection maps AEFG onto AE'F'G?<br><br> HELPPP PLEASE❗️❗️❗️‼️‼️

attashe74 [19]

Answer:the answer is app u welcome text me back if i got it wrong armstrongtamarion

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP FAST I'LL MARK YOU AS BRAINLIST

konstantin123 [22]

Answer: 400

Step-by-step explanation: cube root 8000 then that times itself

5 0

2 years ago