Does anyone know how to do part b?

I am stuck on these last four questions please help if possible.

zloy xaker [14]

Answer:

Step-by-step explanation:

9). -2.2(4 - 1.9x) = 3.3(0.2x - 0.8)

-8.8 + 2.2(1.9x) = 0.66x - 2.64

-8.8 + 4.18x = 0.66x - 2.64

4.18x - 0.66x = 8.8 - 2.64

3.52x = 6.16

x = 1.75

10). 3.2(1 + 2.6x) = 2.4(x - 3.6)

3.2 + 8.32x = 2.4x - 8.64

8.32x - 2.4x = -3.2 - 8.64

5.92x = -11.84

x = 2

11). 4.6(2x - 5.5) = 3.9 + 0.8(1 + 5.5x)

9.2x - 25.3 = 3.9 + 0.8 + 4.4x

9.2x - 25.3 = 4.7 + 4.4x

9.2x - 4.4x = 25.3 + 4.7

4.8x = 30

x = 6.25

12). 0.2(3x + 2.5) - 4.9 = 3.8 - 2.2(x - 5.5)

0.6x + 0.5 - 4.9 = 3.8 - 2.2x + 12.10

0.6x - 4.4 = -2.2x + 15.90

0.6x + 2.2x = 15.90 + 4.4

2.8x = 20.30

x = 7.25

4 0

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.

5 0

3 years ago

Simplify....... -9-6c-j

marin [14]

Answer: −6c−j−9

Step-by-step explanation:

3 0

3 years ago

3. Find two numbers whose product is -6 and whose sum is 1.

Anestetic [448]

Answer:

-2, 3

Step-by-step explanation:

yeah that's it I don't have any explanation or formula

5 0

3 years ago

Read 2 more answers

A survey of 500 likely voters showed that 385 felt that the economy was the most important national issue. Find a point estimate

Dovator [93]

Answer:

Option B

Step-by-step explanation:

Given that a survey of 500 likely voters showed that 385 felt that the economy was the most important national issue.

Sample size n = 500

favor who feel the ecomomy is the most important national issue x= 385

Sample proportion = $\frac{x}{n} \\=0.77$

Sample proportion would be the point estimate for population proportion of voters who feel the ecomomy is the most important national issue.

Hence the point estimate (p-hat0 for p, the population proportion of voters who feel the ecomomy is the most important national issue

is 0.77

(option B)

4 0

3 years ago