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Burka [1]
3 years ago
11

Antonio completed the right column of the table to help him find the sum of 2/9 and 1/5.

Mathematics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

Step 1 is the first error

Step-by-step explanation:

Given

\frac{2}{9} + \frac{1}{5}

Required

Which step contains the first error?

From the attachment, the first step is the first error.

This is so because the fractions can not be written as a fraction with a denominator of 14

Solving further:

\frac{2}{9} + \frac{1}{5}

Find Common Denominator 45

\frac{2*5}{9*5} + \frac{1*9}{5*9}

\frac{10}{45} + \frac{9}{45}

Add:

\frac{10 + 9}{45}

\frac{19}{45}

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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
10) Solve using any method<br> 10x+2y=42<br> 5x+y=20
KIM [24]

Answer:

no solutions

Step-by-step explanation:

10x+2y=42

5x+y=20

Multiply the second equation by -2 to use elimination

-2(5x+y)=20*-2

-10x -2y = -40

Add this to the first equation

10x+2y=42

-10x -2y = -40

--------------------------

0 = 2

This is never true.  This means there are no solutions

3 0
3 years ago
Read 2 more answers
Juan went out to lunch with five friends. They decided to divide the bill evenly among them. Juan estimated that his lunch cost
Lina20 [59]
Juan’s lunch would have costed less if he paid separately. his lunch is 5.25
the bill 37.20
37.20 / 5
$7.44 each kid
7.44 - 5.25 = 2.19
$2.19 less if he paid separately
4 0
3 years ago
Read 2 more answers
Solve for "y" <br> 8y-4x=2
Marizza181 [45]
        -4x + 8y = 2
-4x + 4x + 8y = 2 + 4x
                 8y = 4x + 2
                  8         8
                   y = ¹/₂x + ¹/₄
7 0
3 years ago
6(12 - 5) &gt; 50<br> true or false
Alexxandr [17]
False pls mark meh braliest
5 0
2 years ago
Read 2 more answers
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