What is the measure of angle SRW

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

The total amount of water a tank can hold is 14 1/2 gallons. Raul wants to find out how many 1 1/4 gallon buckets of water can b

sukhopar [10]

Answer: $\frac{29}{2}$ ÷ $\frac{5}{4}$ , $\frac{29}{2}$ × $\frac{4}{5}$

Step-by-step explanation: Let me know if you need an explanation.

8 0

3 years ago

Read 2 more answers

The perimeter of a right triangle is 60 feet and the area is 150 square feet. If the lengths of the sides of the triangle are ea

JulsSmile [24]

Answer:

B. 600 ft

Step-by-step explanation:

The formula for the Perimeter of a right triangle = Side a + Side b + Side c

The perimeter of a right triangle is 60 feet

If the lengths of the sides of the triangle are each multiplied by 10, what is the perimeter of the new right triangle?

Hence, the perimeter of the new right triangle = 10 × 60ft

= 600ft

Option B is the correct Option.

4 0

3 years ago

Read 2 more answers

A polling organization asked a nation-wide random sample of 1600 new brides (those who got married in the past 2 years) the foll

iren [92.7K]

Answer:

It has millions of tickets. On each ticket is written a number a dollar amount. The exact average and SD are unknown but are estimated from the sample to be $20,000 and $5,000 respectively.

Step-by-step explanation:

Given that:

sample size n = 1600

sample mean $\overline x$ = 20000

standard deviation = 5000

The objective is to choose from the given option about what most closely resembles the relevant box model.

The correct answer is:

It has millions of tickets. On each ticket is written a number a dollar amount. The exact average and SD are unknown but are estimated from the sample to be $20,000 and $5,000 respectively.

However, if draws are made without replacement, the best estimate of the average amount for the bride will be $20,000

Similarly, the standard error for the sample mean = $\dfrac{s}{\sqrt{n}}$