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.
We need a picture of the shape.
For this case we have that the area of a circle is given by:
Where:
r: It is the radius of the circle
By clearing the radio we have:
We take the positive value:
The diameter is twice the radius:
Answer:
Answer:
#5 x=8, #6 x=5
Step-by-step explanation:
Theyre vertical so for number 5 you would have the equation
5x+15=55
-15 -15 from both sides
5x=40
divide by 5 from both sides and you get
x=8
Same thing vertical so you would have
16x+7=87
-7 -7 from both sides
16x=80
divide by 16 from both sides
x=5
Answer:
The $1 belongs to the cash box
Step-by-step explanation:
Given
See attachment for complete question
Required
Determine if the $1 belongs to the cash box or not
Represent singles with s and couples with c.
From the attachment, we have:
--- total attendance
--- ticket sold
Solve for s and c.
Make s the subject in (1)
Substitute 47 - c for s in (2)
Open bracket
This means that the total individual which makes up the couples are 35. This is not possible because couples are in 2's and the total should be an even number.
<em>So, we can conclude that the $1 belongs to the cash box</em>
