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3 years ago

Solve 1/2n + 3 < 5 ??

Mathematics

2 answers:

mars1129 [50]3 years ago

8 0

Answer:

n<4

Step-by-step explanation:

WITCHER [35]3 years ago

6 0

Answer:

$n < 4$

Step-by-step explanation:

$\frac{1}{2} n + 3 < 5 \\ subtract \: 3 \: from \: both \: sides \\ \\ \frac{1}{2} n < 2 \\ divide \: both \: sides \: by \: \frac{1}{2} \\ \\ n < 4$

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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

Which expressions are equivalent to 2 (a + 3b)?

VARVARA [1.3K]

Answer:

A & B

Step-by-step explanation:

A is spot-on because of the use of distributive property. B by the use of like terms.

8 0

3 years ago

Which number equals (3)^-3? A) -27 B) -9 C) -1/27 D) 1/27

Alecsey [184]

Answer:

the answer is D 1/27

Step-by-step explanation:

7 0

3 years ago

Select an expression equal to 6(4q + 8r)

Masteriza [31]

Answer: B: 24q+48r

Step-by-step explanation: 6(4q+8r)

=(6)(4q+8r)

=(6)(4q)+(6)(8r)

=24q+48r

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3 years ago

The decibel level of sound in a quiet office is 10−6watts/m2.

Liono4ka [1.6K]

I don’t understand your question can you explain it more

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4 years ago