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3 years ago

An engineer, in an attempt to make a quick measurement, walks 100 ft from the base of an overpass

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

The distance of the overpass above the ground is approximately 26.795 ft

Step-by-step explanation:

The parameters given are;

The distance from the overpass the engineer stands before determining the angle of elevation of the overpass from his standing point = 100 ft

The angle of elevation of the overpass as determined by the engineer from 100 ft = 15°

By trigonometric ratios, we have;

$Tan(\theta) = \dfrac{Opposite \, side \, to\ angle}{Adjacent\, side \, to\, angle}$

The opposite side to the 15° angle of elevation in the above case is the distance of the overpass above the ground

The opposite side to the 15° is the distance of the engineer from the base of the overpass

Therefore;

Tan(15°) the height of the overpass=

length

$Tan(15 ^{\circ}) = \dfrac{The \ distance \, of \, the \ overpass \ above \ ground}{100 \ ft}$

The distance of the overpass above the ground = 100 × tan (15°) ≈ 26.795 ft.

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Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

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slavikrds [6]

Answer:

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Step-by-step explanation:

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emmasim [6.3K]

Answer:

Step-by-step explanation:

If every 175th person gets free admission, that means

175th

175 * 2 = 350th

175 * 3 = 525th

person gets free admission and so on...

This basically means "how many 175s are there is 6742 (for first day) and 5487 (for 2nd day)?"

We would need to divide and see the WHOLE number, not the fractional(decimal) amount remaining. So

1st Day:

6742/175 = 38.53

This means 38 people will get free admission

2nd Day:

5487/175 = 31.35

This means 31 people will get free admission

In the 2 days, total of 38 + 31 = 69 people will get free admission

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Answer:

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Step-by-step explanation:

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castortr0y [4]

-9-(-7)
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