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3 years ago

Explain how to use the regression calculator to make a reasonable prediction.

2 answers:

8 0

Given bivariate data, first determine which is the independent variable, x, and which is the dependent variable, y. Enter the data pairs into the regression calculator. Substitute the value for one variable into the equation for the regression line produced by the calculator, and then predict the value of the other variable.

denis-greek [22]3 years ago

8 0

Answer:

Sample Answer: Given bivariate data, first determine which is the independent variable, x, and which is the dependent variable, y. Enter the data pairs into the regression calculator. Substitute the value for one variable into the equation for the regression line produced by the calculator, and then predict the value of the other variable.

Step-by-step explanation:

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Learning Theory In a typing class,the averege number N of words per minutes typed after t weeks of lessons can be modeled by N =

Jet001 [13]

Answer:

a) $N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

b) $N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

c) $70 =\frac{95}{1+8.5 e^{-0.12t}}$

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

d) If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

Step-by-step explanation:

For this case we have the following expression for the average number of words per minutes typed adter t weeks:

$N(t) = \frac{95}{1+8.5 e^{-0.12t}}$

Part a

For this case we just need to replace the value of t=10 in order to see what we got:

$N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

So the number of words per minute typed after 10 weeks are approximately 27.

Part b

For this case we just need to replace the value of t=20 in order to see what we got:

$N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

So the number of words per minute typed after 20 weeks are approximately 54.

Part c

For this case we want to solve the following equation:

$70 =\frac{95}{1+8.5 e^{-0.12t}}$

And we can rewrite this expression like this:

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

Now we can divide both sides by 8.5 and we got:

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

And then if we solve for t we got:

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

And we can see this on the plot 1 attached.

Part d

If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

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Step-by-step explanation:

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