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iren [92.7K]
3 years ago
10

Extremely Easy Math!!! 50 Points!!! BRAINLIEST!!!

Mathematics
1 answer:
N76 [4]3 years ago
4 0

Answer:

x = 40

Step-by-step explanation:

If you add the total of 3 angles in a STRAIGHT line, you have to get 180 in total.

So,        x =  180 - ( 80 + 60 )

               =  180 - 140

               =  40

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Factor completely x3 − 2x2 − 5x + 10.
Akimi4 [234]

Answer:

(x-2) (x^2 -5)

Step-by-step explanation:

x^3 − 2x^2 − 5x + 10

We will factor by grouping

Take and x^2 out of the first two terms and  -5 out of the last two terms

x^2 (x-2) -5(x-2)

Now we can factor out (x-2)

(x-2) (x^2 -5)

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
Marilyn bought 6 cans of tuna at $1.29 a can. How much did she spend
nika2105 [10]
1.29*6=7.74 so she spend $7.74
3 0
3 years ago
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What is the surface area? 5 in 4 in 2 in
Dennis_Churaev [7]

I think the Answer is 76 in^2

Hope this helps

8 0
3 years ago
The graphs of quadrilateral LMNO and its image are shown below.
Licemer1 [7]

Answer:

6

Step-by-step explanation:

8 0
3 years ago
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