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3 years ago

Do alien raviolis exist?

Mathematics

1 answer:

emmasim [6.3K]3 years ago

4 0

I would assume not. Is this related to school or no? I'm a little lost here lol

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PLEASEEEEE HELPPPP MEEEEEEE BRAINLIEST TO THE FIRST ANSWER AND 30 POINTS

icang [17]

Answer:

CD ≠ EF

Step-by-step explanation:

Using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = C(- 2, 5) and (x₂, y₂ ) = D(- 1, 1)

CD = $\sqrt{(-1+2)^2+(1-5)^2}$

= $\sqrt{1^2+(-4)^2}$

= $\sqrt{1+16}$ = $\sqrt{17}$

Repeat using (x₁, y₁ ) = E(- 4, - 3) and (x₂, y₂ ) = F(- 1, - 1)

EF = $\sqrt{(- 1+4)^2+(-1+3)^2}$

= $\sqrt{3^2+2^2}$

= $\sqrt{9+4}$ = $\sqrt{13}$

Since $\sqrt{17}$ ≈ $\sqrt{13}$ , then CD and EF are not congruent

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

PLEASE HELP!! if f(x)= 2x^2-11, find f(4)

Lemur [1.5K]

Answer: f(4) = 21

Step-by-step explanation:

2(4)^2-11

4^2 = 16

2(16)-11

Multiply 2 and 16

32 - 11

Subtract

4 0

2 years ago

Read 2 more answers

What is the value of Negative 5 Superscript 6? –15,625 –30 30 15,625

babymother [125]

Answer:

15625

Step-by-step explanation:

Lets use the information to present the question into mathematical form:

$(-5)^{6}$

so this means -5 is multiplying itself 6 times so this means

value of negative 5 superscript 6 means

$(-5)^{6}$ = -5 x -5 x -5 x -5 x -5 x -5

so the answer is 15625

7 0

2 years ago

Read 2 more answers

Solve for x.

inessss [21]

Answer:

Step-by-step explanation:

6 0

3 years ago