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3 years ago

What is the slope formula in Geometry? This is for my math homework.

Mathematics

2 answers:

Grace [21]3 years ago

4 0

7 if its addition and 10 if its muiltiplacation

Oxana [17]3 years ago

3 0

The answer for this is 10

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%5Calpha-1%5Cfrac%7B1%7D%7B5%7D%3D2%5Cfrac%7B3%7D%7B4%7D" id="TexFormula1

velikii [3]

I DONT KNOW LOL PLS STOP CHEATING

5 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

Please help me out with this question please!!!

bixtya [17]

Answer:

3 days

Step-by-step explanation:

company A would have the equation 50x+40 =y and company B would be 30x + 100 =y. we're tying to find the days that these cost the same so you would do 50x+40 = 30+100. When you isolate x you get 3

7 0

3 years ago

natka813 [3]

Answer:

x= -8 , y = 5

x= 25/4 , y = 1/4

Step-by-step explanation:

substitute first eqn into the second eqn:

(7 - 3y)^2 -y^2 = 39

49 - 42y + 9y^2 - y^2 = 39

8y^2 - 42y +10 =0

4y^2 - 21y + 5 = 0

(4y-1) (y-5) = 0

y= 1/4 , 5

when y=1/4

x = 7- 3/4

=25/4

when y= 5

x = 7- 15

= -8

5 0

4 years ago

Read 2 more answers

Question is attached

Arlecino [84]

Answer:

Let v = ml of 100% vinegar

Then 150-v = ml of dressing

v + .05(150-v) = .24(150)

v + 7.5 - .05v = 36

.95v = 28.5

v = 28.5/.95

v = 30 ml of vinegar

dressing = 150-30 = 120 ml

Step-by-step explanation:

7 0

3 years ago