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yKpoI14uk [10]
2 years ago
8

HELP ASAP!!!!!!! I NEED THIS SOON!!!

Mathematics
2 answers:
Arte-miy333 [17]2 years ago
7 0

Answer:

Is this a question about the laws of angle of incidence or is it just coordinates

Step-by-step explanation:

Rama09 [41]2 years ago
3 0

Answer:

y - 1 = 0 45 degreees 1 90 degrees x times 3y

Step-by-step explanation:

Trust

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Find X when x equals negative 15
Ann [662]
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3 years ago
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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3 years ago
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gtnhenbr [62]

Answer:

I Believe you are correct

Step-by-step explanation:

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3 years ago
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