Simplify ( Write without the absolute value sign) |x-(-12)|, if x> -120

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

Erin has 5 cans of juice and drinks 13 of a can daily. In how many days will she empty all the cans she has?

professor190 [17]

Answer:

15 days

Step-by-step explanation:

Don't you mean 1/3? If so,

5 cans

-------------------- = 15 days

1/3 can/daily

5 0

2 years ago

HELPPPP!!!

Hatshy [7]

Answer:

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

(x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

∴ (x² - 2x) - 4(y² - 4y) - 31 = 0

- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵ (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

the bracket to keep the equation as it

∴ 4[y² - 4y + 4) - 4]

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify

∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms

∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides

∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - (y - 2)²/4 = 1

∵ 16 = (4)² and 4 = (2)²

∴ The standard form of the equation of the hyperbola is

(x - 1)²/4² - (y - 2)²/2² = 1

4 0

3 years ago

Add. Write your answer in simplest form.

alexandr1967 [171]

Answer:

yes, you got it correct

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Hot dogs are sold in packages of 8, and hot dog buns are sold in packages of 12. What's the smallest number of each that could b

PIT_PIT [208]

24 is the answer 8x3 = 24 12x2 = 24 8*3 = 12*2

7 0

3 years ago