(8n-1) +3(3n+2) solve

Out of 400 applicants for a job, 179 are female and 72 are female and have a graduate degree. Step 2 of 2 : If 118 of the applic

Dafna11 [192]

Answer:

36/59 or 0.610

Step-by-step explanation:

P(female and degree)/P(degree)

72/118

36/59

0r 0.610

7 0

3 years ago

Nathaniel bought headphones online for $22. He used a coupon code to get a 40% discount. The website also applied a 10% processi

Tatiana [17]

Answer:

8.80 if it's the 40°/.

2.20 if it's the 10°/. processing fee

13.20 for how much he paid

8 0

2 years ago

Hannah can paint a room in 8 hours. Destiny can paint the same room in 6 hours. How long does it take for both Hannah and Destin

Ksivusya [100]

Answer:

3 hours 26 minutes

Step-by-step explanation:

1/8 + 1/6 = 1/t where t is the required time in hours.

Multiply through by 24t:-

3t + 4t = 24

7t = 24

t = 3 3/7 hours

or 3 hours 26 minutes to the nearest minute.

6 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

Which expressions are equivalent to when x0? Check all that apply.

Genrish500 [490]

We have that

$\frac{(x+4)}{3} / \frac{6}{x} = \frac{x*(x+4)}{3*6} \\ \\ = \frac{( x^{2} +4x)}{18}$

therefore

case a)
$\frac{(x+4)}{3} * \frac{x}{6}$
Is equivalent

case b)
$\frac{6}{x} * \frac{(x+4)}{3}$
Is not equivalent

case c)
$\frac{x}{6} * \frac{(x+4)}{3}$
Is equivalent

case d)
$\frac{(2 x^{2} +4x)}{6}$
Is not equivalent

case e)
$\frac{(2 x^{2} +4x)}{18}$
Is equivalent

Hence

the answer is

$\frac{(x+4)}{3} * \frac{x}{6}$

$\frac{x}{6} * \frac{(x+4)}{3}$

$\frac{(2 x^{2} +4x)}{18}$

3 0

3 years ago

Read 2 more answers