F(a)=a^2=4a;find f(-2)

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

1 tenth is how many times greater than 1 hundredth

Anna35 [415]

Answer:

10

Step-by-step explanation:

1 tenth: 1/10

1 hundredth: 1/100

1/10 = X × 1/100

X = 100/10

X = 10

7 0

3 years ago

Read 2 more answers

Hilda draws a card from a deck of cards and it is a seven. The probability that it is the 7 of spades is Select one: O a. 1/3 O

Schach [20]

Answer:b

Step-by-step explanation:

Given

Hilda draws a card from a deck of cards and it is seven.

It is certain that card is 7 so there a 4 cards which can be 7

i.e. 7 of hearts,7 of spades,7 of club,7 of diamond

So probability that it is the 7 of spades is

$\frac{1}{4}=0.25$

8 0

3 years ago

The graph of function f is shown. Function g is represented by the table. Which statement correctly compares the two functions?

Dafna1 [17]

Answer:

Option (A)

Step-by-step explanation:

Graph of function 'f' represents,

x - intercept of the function 'f' → (1, 0)

y - intercept of the function → (0, 6)

As x-approaches ∞, value of the function approaches (-2)

Points in the given table is for the another function 'g'

x - intercept of the function 'g' → (1, 0) [For x - intercept, y = 0]

y - intercept of the function 'g' → (0, 3) [For y - intercept, x = 0]

As x approaches ∞, value of function 'g' approaches (-1).

Therefore, x - intercepts of both the functions are same but end behavior are different when x → ∞.

Option (A) will be the answer.

7 0

2 years ago

Write a linear equation in point slope form that passes through the point (-2,5) with a slope or 1/2

trasher [3.6K]

Answer:

$y-5=\frac{1}{2}(x+2)$

Step-by-step explanation:

Point Slope is:

$y-y_1=m(x-x_1)$

We are given the point (-2, 5) and a slope of 1/2.

$y-y_1=m(x-x_1)\rightarrow\boxed{y-5=\frac{1}{2}(x+2)}$

Hope this helps.

4 0

2 years ago