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tatuchka [14]
3 years ago
15

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:

12

7

2

- 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
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