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erica [24]
4 years ago
9

Develop a three-month moving average for this time series. Compute MSE and a forecast for month 8. If required, round your answe

rs to two decimal places. Do not round intermediate calculation.
Engineering
1 answer:
Trava [24]4 years ago
4 0

Answer:

The missing Table is;

Month  1  2  3  4  5  6  7

Value  24  13  21  14  20  23  15

MSE = 20.56

Forecast = 19.33

Explanation:

<u>Table Formation</u>

Month Value Forecast Error  Absolute Error Error Square

1          24    

2          13    

3          21    

4          14         19.33333 -5.33333 5.333333 28.44444

5          20           16                       4              4             16

6          23          18.33333 4.666667 4.666667 21.77778

7          15           19                   -4                       4             16

   

    20.55

    19.33

<u>Table Explanation</u>

In the above table, we first calculate the forecast column. A forecast is taken as the sum of the three entries in order and then taking out the average.

Error is calculated by subtracting the Actual value from the forecast value.

Then we take the absolute of the forecast value.

Then we take the square of the absolute of the forecasted values.

Now we calculate MSE as;

MSE = ( 28.44 + 16 + 21.78 + 16 ) / 4

<em>MSE = 20.56</em>

Forecast for 8th month  = ( 20 + 23 + 15 ) / 3

Forecast for 8th month  = 19.33

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Answer with Explanation:

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We know that for solid shaft I_{p}=\frac{\pi R^4}{2}

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Since the two shafts are subjected to same torque from the relation of Torque we have

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2) For hollow shaft we have

\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}

Comparing the above 2 relations we see

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Part b)

Strength of solid shaft = \tau _{max}=\frac{T\times R}{I_{solid}}

Weight of solid shaft =\rho \times \pi R^2\times L

Strength per unit weight of solid shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}

Strength of hollow shaft = \tau '_{max}=\frac{T\times R}{I_{hollow}}

Weight of hollow shaft =\rho \times \pi (R^2-0.7R^2)\times L

Strength per unit weight of hollow shaft = \frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}

Thus \frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16

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Answer:

Following are the proving to this question:

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