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aksik [14]
3 years ago
7

A road crosses a railroad track at the top of a steep hill.The train cannot stop for oncoming cars, and cars cannot see the trai

n until it is too late. Suppose a train begins crossing the road at time t1 and that the car begins crossing the track at time t2, where 0 < t1 < T and 0 < t2 < T
Required:
a. Find the sample space of this experiment.
b. Suppose that it takes the train al seconds to cross the road and it takes the car a2 seconds to cross the track. Find the set that corresponds to a collision taking place.
c. Find the set that corresponds to a collision is missed by 1 second or less.
Mathematics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

a. Sample space = { (t₁ , t₂ ) | 0 < t₁ < T , 0 < t₂ < T }

b.A₁ = { t₂ < t₁ < t₂ + a₂ }

  A₂ = { t₁ < t₂ < t₁ + a₁ }

  if t₁ - t₂ < a₂ , t₂ - t₁ < a₁

c. A₃ = { t₂ + a₂ < t₁ < t₂ + a₂ + 1 }

   A₄ = { t₁ + a₁ < t₂ < t₁ + a₁ + 1 }

Step-by-step explanation:

As given,

A train begins crossing the road at time t₁

The car begins crossing the track at time t₂

where 0 < t₁ < T and 0 < t₂ < T

a.)

Sample space = { (t₁ , t₂ ) | 0 < t₁ < T , 0 < t₂ < T }

b.)

Given that -

The train takes a₁ seconds to cross the road

The car takes a₂ seconds to cross the track.

So, The set that correspond to collision is

A₁ = { t₂ < t₁ < t₂ + a₂ }

A₂ = { t₁ < t₂ < t₁ + a₁ }

if t₁ - t₂ < a₂ , t₂ - t₁ < a₁

c)

The set that corresponds to a collision is missed by 1 second or less is -

A₃ = { t₂ + a₂ < t₁ < t₂ + a₂ + 1 }

A₄ = { t₁ + a₁ < t₂ < t₁ + a₁ + 1 }

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A special type of autonomous differential equation, called the logistic equation, is typically written as Cape = rP(1- ), and is
kipiarov [429]

Answer:

a) dP / dt = 0.15*P ( 1  - P/750)   ,  r = 0.15 , K = 750

b) decreasing : ( - inf , 0 ) & ( 750 , inf )

    increasing : ( 0 , 750 )

Step-by-step explanation:

Given:

- The standard for of logistic equation:

                       dP / dt = r*P( 1 - P/K)

Where, r and K are constants.

- The given experimental relation is:

                       dP / dt = 0.15*P - 0.0002*P^2

Find:

Rewrite the equation in the typical form to determine the intrinsic growth rate(r) and environmental carrying capacity (K).

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Solution:

- First we will convert the given relation into standard form as follows:

                         dP / dt = 0.15*P - 0.0002*P^2

Factor out 0.15*P:

                         dP / dt = 0.15*P ( 1  - P/750)

- Hence, our constants are:

                         r = 0.15 , K = 750

- We will set up and inequality for what values of P is dP/dt > 0 and dP/dt <0

                          dP / dt = 0.15*P - 0.0002*P^2 < 0

= Solve for P:

                          P < 0 , 1-P/750 = 0 -----> P > 750

So when when P is decreasing the intervals are:

                          ( - inf , 0 ) & ( 750 , inf )

And when P is increasing the intervals are:

                              ( 0 , 750)

         

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3 years ago
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Answer:

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6 0
2 years ago
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yarga [219]

Answer:

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Step-by-step explanation:

The given polynomial is: 3x³ - 24x² + 45x

To factorize it: 3x³ - 24x² + 45x

Since, 3x is common through out we can take it out.

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Now we have to factorize (x² - 8x + 15)

The roots of this quadratic equation is: 5, 3.

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Answer:

840

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3 years ago
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