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OleMash [197]
3 years ago
8

13. Given the following historical data and weights of .5, .3, and .2, what is the three-period weighted moving average forecast

for period 5? Period Value 1 138 2 142 3 148 4 144 a. 144.2 b. 144.8 c. 144.6 d. 143 e. 144
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

b. 144.8

Step-by-step explanation:

When calculating the moving average estimate of an observation , each of the observations are usually computed with the same weighted . In some cases, it is  beneficial to assign different  weight on the observations  such that the observation closer to the time period being forecast, has higher weight.  This is refer to as weighted moving average technique. The sum of the individual weight in a weighted moving average technique must equal to 1.

The three-period weighted moving average forecast for period 5 = 144*0.5 + 148 *0.3 + 142 *0.2 =  144.8

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Part A) The area of the figure is 24\ units^{2}

Part B) The perimeter of the figure is 20\ units

Step-by-step explanation:

step 1

Find the area of the figure

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A=\frac{1}{2}bh

<u>Area of triangle ABD</u>

Observing the graph

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substitute

A=\frac{1}{2}(6)(4)=12\ units^{2}

<u>Area of triangle BCD</u>

Observing the graph

b=BD=(-2+8)=6\ units

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A=\frac{1}{2}(6)(4)=12\ units^{2}

The area of the figure is

12\ units^{2}+12\ units^{2}=24\ units^{2}

step 2

Find the perimeter of the figure

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The perimeter of the figure is equal to

P=AB+BC+CD+AD

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the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units

Find the distance BC

d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units

Find the distance CD

d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units

Find the distance AD

d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units

substitute the values

P=5+5+5+5=20\ units

7 0
3 years ago
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