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Bogdan [553]
3 years ago
7

John assignments in value of his car overtime the equation for the line of best fit is approximate as Y represents 2.9 X +17.7 w

here y represents the value in thousands of dollars what values complete the Residual table
Mathematics
1 answer:
ycow [4]3 years ago
5 0

Answer:

For the given equation of the line of best fit, the values that complete the table are as follows

An equation for a line of best fit follows the same guidelines as a linear function where 'y' represents the total (value), 'x' represents time (years), -2.9 is the rate, and 17.7 is the starting value.  The table indicates that for any year, there is a given value, but what we are solving for is the predicted value.  The residual is the different between the given and predicted values.  So, for 'a', we need to solve for the 'y' in our equation by replacing 'x' with '1', multiplying by -2.9 and adding 17.7.  This gives us 14.8.  For 'b', we simply need to subtract the given and predicted values to get a residual of 0.1.  For 'c', we again solve for 'y' by replacing 'x' with '3' in our given equation to get 9. And, for 'd' we subtract the given value of 5 and the predicted value of 6.1 to get 1.1.

answer-

a = 14.8

b = 0.1

c = 9

d = -1.1

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Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

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S5 = 0.6333

=========================================================

Explanation:

Let f(n) = \frac{(-1)^{n+1}}{n!}

The summation given to us represents the following

\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\

There are infinitely many terms to be added.

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

The partial sums only care about adding a finite amount of terms.

The partial sum S_1 is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say S_1 = f(1) = 1

I'm skipping the steps to compute f(1) since you already have done so.

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

The second partial sum is when things get a bit more interesting.

We add the first two terms.

S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\

The scratch work for computing f(2) is shown in the diagram below.

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

We do the same type of steps for the third partial sum.

S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\

The scratch work for computing f(3) is shown in the diagram below.

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

Now add the first four terms to get the fourth partial sum.

S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like S_3 = S_2 + f(3) and S_4 = S_3 + f(4)

In general, S_{n+1} = S_{n} + f(n+1) so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

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

Let's use that recursive trick to find S_5

S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333

The scratch work for f(5) is shown below.

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