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.
2 1/4 = 2 2/8. So, simply subtract 7 7/8 - 2 2/8 to get your final answer of 5 5/8
20 seconds faster than Jesse so one minute and 57 seconds is my best guess.
Answer:
Explanation:
The formula for calculating the distance between two points is:
d
=
√
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
Substituting the values from the points in the problem gives:
d
=
√
(
−
6
−
−
3
)
2
+
(
−
4
−
−
5
)
2
d
=
√
(
−
6
+
3
)
2
+
(
−
4
+
5
)
2
d
=
√
(
−
3
)
2+
1
2
d
=
√
9+
1
d
=
√
10
Or
d
=
3.162
rounded to the nearest thousandth
