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3 years ago

While hovering near the top of Ribbon Falls in Yosemite National Park at 1600 feet, a

Mathematics

1 answer:

ella [17]3 years ago

4 0

Answer:

a)The sunglasses hit the ground at t=0 seconds

b)The maximum height h(t) = 1600feet

Step-by-step explanation:

Step(i):-

Given that the polynomial function

h(t) = -16 t² + 1600 ...(i)

Differentiating equation (i) with respective to 't' , we get

$\frac{dh}{dt} = -16(2t)+0$

Equating zero

$\frac{dh}{dt} = 0$

⇒ -32t =0

⇒ t =0

The sunglasses hit the ground at t=0 seconds

Step(ii):-

Again differentiating with respective to 't', we get

$\frac{d^{2} h}{dt^{2} } = -32(1) = -32$

The maximum height at t=0

h(t) = -16 t² + 1600

h(t) = 1600feet

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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.

5 0

3 years ago

4x+3=x+9 What is the number? X=?

Aneli [31]

Answer:

Step-by-step explanation:

4x+3=x+9

4x-x=9-3

3x=6

x=6/3

x=2

8 0

2 years ago

Read 2 more answers

The perimeter of a notebook is 44 inches the notebook is 10 inches wide how tall is it?

tatyana61 [14]

Answer=36inches tall

Explanation= 44+10= 54
90-54=36in

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3 years ago

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Fudgin [204]

Hey mate hope its help you...

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3 years ago

Read 2 more answers

chelsea takes the bus to school,but she walks home. the bus travels to the school at an average speed of 45 miles per hour. whil

jekas [21]

For this case, the first thing you should know is:
d: v * t
Where,
d: distance
v: speed
t: time
To go to school by bus we have:
d = 45 * t
To return from school we have:
d = 2.5 * (1-t)
how the distance is the same:45 * t = 2.5 * (1-t)
Answer:
the equation that will help find the time it takes chelsea to get to school on the bus is:
45 * t = 2.5 * (1-t)
Note: the equation will have one solution.

4 0

3 years ago