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sladkih [1.3K]
2 years ago
9

- 1 (multiplicity 3), 3 (multiplicity 4)

Mathematics
1 answer:
Vinil7 [7]2 years ago
4 0

Answer:

-1,81

Step-by-step explanation:

Multiplicity of a number can be written as power

for example

-1(multiplicity 3) can be written as

-1^{3}= -1

similiarly 3(multiplicity 4) can be written as

3^{4}=81


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A railway engine is of mass 72tonnes and 11m long. An exact scale model is made of it and is 44cm long. find the mass of the mod
34kurt

Answer:

it is correct

Step-by-step explanation:

2000 N

The resistance force acting is 1N per ton.

Tension in the coupling between engine and the wagon is 2000N.

4 0
2 years ago
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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
Help please and thank you!!!
Ghella [55]

Answer:

The slop is undifined cause ur basicly just falling cause theirs no x just straight down forever


8 0
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WILL GIVE BRAINLIEST AND 30 POINTS<br><br> What are 2 expressions that are equivalent to 5(2j+3+j)
VMariaS [17]
5(2j+3+j)
= 5(2j) + 5(3) + 5(j) we distribute 5 to all of the expression
= 10j + 15 + 5j connect like terms j
= 15j + 15
6 0
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Read 2 more answers
What is 18 divided by negative 3
sweet-ann [11.9K]
18 / -3 = -6

answer

<span>negative 6 is 18 divided by negative 3
</span>
hope it helps
6 0
3 years ago
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