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3 years ago

A regular pyramid has a_

Mathematics

2 answers:

nydimaria [60]3 years ago

8 0

Answer is :C) regular polygon

ankoles [38]3 years ago

8 0

I think the answer is C.

You might be interested in

Russell works at a football concession stand selling drinks for $2 and bags of chips for $3

Reptile [31]

Answer:

its 5

Step-by-step explanation:

2+3 = 5

-_-

4 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

Can someone tell me the formula i would use for this and how to solve through?

timama [110]

Since they are congruent, cpctc,
just set the congruent parts equal to each other.
so 2x-20=30
and
15=3y-9

3 0

3 years ago

PLEASE HELP WILL CROWNNNNNN!!!!!

Serga [27]

Answer:

- 3

- 8

Step-by-step explanation:

f ( - 2 )

= 2 - 5 ( - 2 )

= 2 - ( - 10 )

= 2 + 10

= 12

f ( - 1 )

= 2 - 5 ( - 1 )

= 2 - ( - 5 )

= 2 + 5

= 7

f ( 0 )

= 2 - 5 ( 0 )

= 2

f ( 1 )

= 2 - 5 ( 1 )

= 2 - 5

= - 5 + 2

= - 3

f ( 2 )

= 2 - 5 ( 2 )

= 2 - 10

= - 10 + 2

= - 8

6 0

3 years ago

If a = 7 and b = 11, what is the measure of ∠B? (round to the nearest tenth of a degree)

Firdavs [7]

Is there a picture for this if so can you post it

8 0

3 years ago