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aniked [119]
3 years ago
13

A train travel 20 miles in 15 minutes 40 miles in 30 minutes and 60 miles in 45 minutes which equation represents the speed of t

he train in miles per hour
Mathematics
2 answers:
Verizon [17]3 years ago
6 0

Answer:

y = 80x

Step-by-step explanation:

In 30 minutes the train went 40 ml so in an hr the train would be at 80 ml

ExtremeBDS [4]3 years ago
4 0
Every 20 miles it went 5 minutes slower
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Galina-37 [17]

Answer:

Step-by-step explanation:

Two angles form a linear pair.  

The measure of one angle is 3.5 times the measure of the other angle.  

Find the measure of each angle.

The measure of one angle is 180º/4.5 or 40º

The measure of the other angle is 140º

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It gives the non negative value of that number irrespective of the sign of the number.

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So in this case;

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Y = -3x + 11<br> y = -6x - 13
Marysya12 [62]

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Problem 7.43 A chemical plant superintendent orders a process readjustment (namely shutdown and setting change) whenever the pH
vaieri [72.5K]

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)

of readjusting (that is, the probability that the measurement is not in the acceptable region) when the process is operating as intended and \mu = 7.0 probability

(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

Answer:

a

The value is  P(X < 6.92 or X > 7.08 ) =  0.26431  

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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