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cricket20 [7]
3 years ago
5

Suppose a student carrying a flu virus returns to an isolated college campus of 6000 students. Determine a differential equation

governing the number of students x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it. (Use k > 0 for the constant of proportionality and x for x(t).)
Mathematics
1 answer:
rodikova [14]3 years ago
6 0

Answer:

\dfrac{dx}{dt}= kx[6000-x], x(0)=0, k>0

Step-by-step explanation:

Total Number of Students =6000

Number of students who have contracted the flu =x(t)

Number of students who have not contracted the flu =6000- x(t)

Now, the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it.

\dfrac{dx(t)}{dt}\propto x(t)[6000-x(t)] \\$Introducing the constant of proportionality, k, we have:\\\dfrac{dx}{dt}= kx[6000-x]

Initially, the campus is uninfected, therefore: x(0)=0

Therefore, a differential equation governing the number of students x(t) who have contracted the flu is:

\dfrac{dx}{dt}= kx[6000-x], x(0)=0, k>0

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3 years ago
Bob hiked 16 miles each day on a 12 day hiking trip. Lola hiked 14 miles each day on her 16 day hiking trip. how many more miles
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Hope this helps!
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3 years ago
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What should you do to solve the equation? 45 = x + 38
antoniya [11.8K]

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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
54) Ashley and Nicole each improved their yards by planting daylilies and ornamental grass. They
Assoli18 [71]

Answer:

Daylilies- $11, Ornamental grass- $7

Step-by-step explanation:

For this one, you want to convert this into simultaneous equations.

If we let d be cost of one daylily and o be cost of one bunch of ornamental grass, we can write the equations as 2d+3o=43 and 6d+3o=87.

From there, we can see they share a like term (3o), so we subtract one equation from the other (since they both are preceded by a +). This gives 4d=44.

Dividing both sides by 4 isolates d, leaving d=$11.

We can substitute this value into either equation to work out o (I'm using the first one as it is easier). This makes it 2×11+3o=43, or 22+3o=43.

From there, we subtract 22 from both sides, giving 3o=21, and dividing by 3 gives a final result of o=$7.

**This content involves writing and solving simultaneous equations, which you may wish to revise. I'm always happy to help!

All working out:

<em>d- cost of one daylily o- cost of one bunch of ornamental grass</em>

<em>2d+3o=43 6d+3o=87</em>

<em />

6d+3o=87

 -     -     -

2d+3o=43

=     =    =

4d      =44

÷4        ÷4

d        =11

2d+3o=43

2×11+3o=43

22+3o=43

3o=21

o=7

3 0
3 years ago
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