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2 years ago

Any suggestions or answers

Mathematics

1 answer:

Bad White [126]2 years ago

7 0

Answer:

3m - 4

Step-by-step explanation:

Divide.

(15mn)/(5n) = 3m

(-20n)/(5n) = -4

3m - 4 is your answer.

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3 part question. If answer all(brainilest will be given.

docker41 [41]

Answer:

1. 1080

2. It shows the initial debt, i.e., $1080

3.24 months

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2 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.

5 0

2 years ago

Simplify. 7/8+(−2/3) divided by 5/6

Sedaia [141]

7/8+(−2/3) divided by 5/6

(7/8 - 2/3)/(5/6)
=(21/24 - 16/24) / (5/6)
= (5/24) / (5/6)
= 5/24 * 6/5
= 6/24
= 1/4

7 0

3 years ago

Read 2 more answers

Happy new year, add 2000 + 21 = 2021 trying to be fun and creative.

uranmaximum [27]

HAPPY NEW YEAR:)

not only today

always be fun and creative

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2 years ago

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write an equation in y=mx+b for a line that passes through (3,-4) & has a slope of -3. you can write it in point-slope and c

dedylja [7]

Y = -3x + b
Plug in point
-4 = -3(3) + b
-4 = -9 + b, b = 5
Solution: y = -3x + 5

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3 years ago