Solve these equations. show on a number line. |x-3|=0

Find the slope for (3,4) and (3,-4)

saveliy_v [14]

Answer:

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The problem:

Find the slope for the line going through (3,4) and (3,-4).

Step-by-step explanation:

Line up points vertically and subtract.

Then put 2nd difference over 1st.

( 3 , 4)

-(3 , -4)

------------

0 8

So the slope would have been 8/0 but this is undefined.

So the slope is undefined.

Also notice the x's are the same and the y's are difference so this is a vertical line. There is only rise in a vertical line and no run. Recall, slope is rise/run. You cannot divide by 0 so this is why we say the slope is undefined when the x's are always the same no matter the y.

7 0

3 years ago

Please help I need someone to help me ASAP. No trolling

nata0808 [166]

Answer:

It will be 16/8 I hope you are happy

6 0

3 years ago

I need help, please.

Usimov [2.4K]

2 rooms would be 62+42 which is 104

3 rooms would be 104+42 which is 146

4 rooms would be 146+42 which is 188

5 rooms would be 188+42 which is 230

6 rooms would be 230+42 which is 272

but these are the prices including the 20$

without the 20$ the prices would be

2 rooms is 84

3 rooms is 126

4 rooms is 168

5 rooms is 210

6 rooms is 252

4 0

3 years ago

Read 2 more answers

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

so

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago

A true-false test consists of 50 questions. how many does a student have to get right to convince you that he is not merely gues

umka2103 [35]

Hmmm. 20 or 30 so the teacher will count the rest,if a student guessed, wrong or right

7 0

3 years ago