All categories

3 years ago

:D PLS HALLLPP ASAP WILL GIVE BRAINLIEST

Mathematics

2 answers:

Papessa [141]3 years ago

5 0

32

I know this Bc I looked it up

Westkost [7]3 years ago

4 0

Answer:

Step-by-step explanation:

brainliest?

You might be interested in

2) This is a line segment between the center and a point on the circle or sphere.

Dima020 [189]

Answer:

Radius

Step-by-step explanation:

A circle is made of specific components which define its shape. It has a fixed center which is equidistant apart from all points at its edge. The distance from the center to each point is called its radius. The distance from its edge through its center to the opposite edge is called the diameter.

3 0

2 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

Need help with two questions please

balandron [24]

Answer:

v is negative for every x, meaning the domain is all values in R, whatever goes under the squared sign will come out as positive and the left out -ve sign will make V negative for all x in R

range of f(x) is -inf to +inf

Step-by-step explanation:

the cubic power doesnt affect the sign

3 0

3 years ago

Please help me with this homework

Vera_Pavlovna [14]

Answer:

linear

non-linear

Step-by-step explanation:

8 0

3 years ago

Can someone pls help me

Alina [70]

Answer:

Sure, what do you need?

4 0

2 years ago