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4 years ago

(-36 4/9) - (-10 2/9) - (18 2/9) solve

Mathematics

1 answer:

PSYCHO15rus [73]4 years ago

6 0

Convert 36 4/9 to an improper fraction. Use this rule: a b/c = ac + b/c

-36 × 9 + 4/9 - (-10 2/9) - 18 2/9

Simplify 36 × 9 to 324

- 324 + 4/9 - (-10 2/9) - 18 2/9

Simplify 324 + 4 to 328

- 328/9 - (-10 2/9) - 18 2/9

Convert 10 2/9 to an improper fraction. Use this rule: a b/c = ac + b/c.

-328/9 - (- 10 × 9 + 2/9) - 18 2/9

Simplify 10 × 9 to 90

-328/9 - (- 90 + 2/9) - 18 2/9

Simplify 90 + 2 to 92

- 328/9 - (-92/9) - 18 2/9

Convert 18 2/9 to an improper fraction. Use this rule: a b/c = ac + b/c.

-328/9 - (- 92/9) - 18 × 9 + 2/9

Simplify 18 × 9 to 162

-328/9 - (- 92/2) - 162 + 2/9

Simplify 162 + 2 to 164

-328/9 - (- 92/9) - 164/9

Simplify brackets

-328/9 + 92/9 - 164/9

Join the denominators

-328 + 92 - 164/9

Simplify

-400/9

Convert to a mixed fraction

$-44\frac{4}{9}$

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Answer:

D. 0.65

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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

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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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