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mel-nik [20]
3 years ago
15

I need help i will give them 100 points.

Mathematics
2 answers:
lilavasa [31]3 years ago
8 0

Hello!

∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘  ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘

<h3>Slope Formula:</h3>

\bold{\displaystyle\frac{y_2-y_1}{x_2-x_1}}

<h3>Where</h3>

\bold{y_2}  is the y-coordinate of the second point

\bold{y_1} is the y-coordinate of the first point

\bold{x_2} is the x-coordinate of the second point

\bold{x_1} is the x-coordinate of the first point

Check the First example:

\bf{y_2} = 6

\bf{y_1}= -2

\bf{x_2}=3

\bf{x_1} = -9

<h3>Plug in the Values</h3>

\bf{\displaystyle\frac{6-(-2)}{3-(-9)} }

6-(-2) is the same as 6+2; same with 3-(-9)

\bf{\displaystyle\frac{6+2}{3+9}}

Simplify:

\bf{\displaystyle\frac{8}{12}}

Is the slope negative? Nope!

<em>How about Option B?</em>

<em />\bf{\displaystyle\frac{7-9}{5-6}}

\bf{\displaystyle\frac{-2}{-1}}

Right now it seems like the slope's negative, huh? Well, wait for it...

\bf{2}

Kaboom!

Please remember that

\bigstar^\circ A negative number multiplied/divided by a negative number equals... a positive! :-)

We have 2 options left: C & D.

\displaystyle\frac{-8-1}{8-(-8)}

\displaystyle\frac{-8-1}{8+8}}

\displaystyle\frac{-9}{16}}

We did it! The slope's actually negative :)

How about Option D (just to make sure the third option is right)

-9-(-8)/-1-2

-9+8/-3

-1/-3

Again we have a negative over a negative.

<h3>Therefore, the correct option is</h3>

\boxed{\boxed{\bold{Option~C}}}

Hope everything is clear.

Let me know if you have any questions!

#LearnWithJoy

\text{An~Emotional~Helper}

\rule{2}{2}~\rule{2}{2}~\rule{2}{2}~\rule{2}{2}~\rule{2}{2}~\rule{2}{2}

Nat2105 [25]3 years ago
5 0
The last one because the y is negative
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Problem 7.43 A chemical plant superintendent orders a process readjustment (namely shutdown and setting change) whenever the pH
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Complete Question

Problem 7.43

   A chemical plant superintendent orders a process readjustment (namely shutdown and setting change) whenever the pH of the final product falls below 6.92 or above 7.08. The sample pH is normally distributed with unknown mu and standard deviation 0.08. Determine the probability:

(a)

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(b)

of readjusting (that is, the probability that the measurement is not in the acceptable region) when the process is slightly off target, namely the mean pH is \mu = 7.02

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a

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b

The value  is P(X < 6.92 or X > 7.08 ) =  0.29344  

Step-by-step explanation:

From the question we are told that

  The mean is  \mu =  7.0

  The standard deviation is  \sigma  =  0.08

Considering question a

Generally the probability of readjusting when the process is operating as intended and mu  7.0 is mathematically represented as

       P(X < 6.92 or X > 7.08 ) =  P(X <  6.92  ) + P(X > 7.08)

=>    P(X < 6.92 or X > 7.08 ) =  P(\frac{X - \mu }{\sigma} < \frac{6.9 - 7}{0.08}  ) + P(\frac{X - \mu}{\sigma} >  \frac{7.08 - 7}{0.08} )

Generally

 \frac{X - \mu }{\sigma} = Z(The \  standardized \  value \ of \  X)

So

=> P(X < 6.92 or X > 7.08 ) =  P(Z < \frac{6.9 - 7}{0.08}  ) + P(Z >  \frac{7.08 - 7}{0.08} )    

=> P(X < 6.92 or X > 7.08 ) =  P(Z < -1.25) + P(Z >  1 )  

From the z table the probability of  (Z < -1.25) and  (Z >  1 ) is  

       P(Z < -1.25) =  0.10565

and

      P(Z >  1 ) = 0.15866

So

=> P(X < 6.92 or X > 7.08 ) =  0.10565 + 0.15866      

=> P(X < 6.92 or X > 7.08 ) =  0.26431      

Considering question b

Generally the probability of readjusting when the process is operating as intended and mu  7.02 is mathematically represented as

       P(X < 6.92 or X > 7.08 ) =  P(X <  6.92  ) + P(X > 7.08)

=>    P(X < 6.92 or X > 7.08 ) =  P(\frac{X - \mu }{\sigma} < \frac{6.9 - 7.02}{0.08}  ) + P(\frac{X - \mu}{\sigma} >  \frac{7.08 - 7.02}{0.08} )

Generally

 \frac{X - \mu }{\sigma} = Z(The \  standardized \  value \ of \  X)

So

=> P(X < 6.92 or X > 7.08 ) =  P(Z < \frac{6.9 - 7.02}{0.08}  ) + P(Z >  \frac{7.08 - 7.02}{0.08} )    

=> P(X < 6.92 or X > 7.08 ) =  P(Z < -1.5) + P(Z >  0.75 )  

From the z table the probability of  (Z < -1.5) and  (Z >  0.75 ) is  

       P(Z < -1.5) = 0.066807

and

      P(Z >  0.75 ) = 0.22663

So

=> P(X < 6.92 or X > 7.08 ) = 0.066807 + 0.22663      

=> P(X < 6.92 or X > 7.08 ) =  0.29344      

6 0
3 years ago
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