Answer:
The diagram of the plotting point
is attached below.
Step-by-step explanation:
Given the points

as


so the point can be visualized as:

Now, we can check the point x = 3.5, and determine the corresponding value y = 2.75 and plot the point at the location (x, y) = (3.5, 2.75)
The diagram of the plotting point
is attached below.
i would say D) 12 is your answer
For a smoothing constant of 0.2
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90
Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12
The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80
Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17
The mean forecast for period 11 is 53.56
Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
Answer: y=Mx+b
Step-by-step explanation:
G
The answer would be -14<span>√2 or the decimal form is 19.79898987...</span>