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1 month ago

A 5 1/2 quart pot is filled 2/3 of the way with water. How many more quarts of water can the pot hold?

Mathematics

2 answers:

Inessa [12.5K]1 month ago

8 0

The pot has the capability to hold 1/3 more than its current level. Therefore, 1/3 of 5 1/2 quarts equals:

(1/3)(11/2) = 11/6 quarts. Thus, the total capacity of the pot is 11/6 quarts.

lawyer [12.5K]1 month ago

3 0

Response:

Water needed to fill the pot $=\frac{11}{6}\texttt{ quarts}$

Detailed explanation:

Capacity of the pot = $5\frac{1}{2} = \frac{11}{2}$ quart

Proportion of the pot currently filled with water = $\frac{2}{3}$

Proportion of the pot that is empty = $\frac{1}{3}$

The additional quarts of water that the pot can accommodate are noted below

Amount of quarts $=\frac{1}{3}\times \frac{11}{2}=\frac{11}{6}\texttt{ quarts}$

Water needed to complete filling the pot $=\frac{11}{6}\texttt{ quarts}$

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Based on the table below listing the costs for the Lenovo zx-81 chip over the past 12 months

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Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
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The forecast for period $F_{t+1}$ is calculated with the formula

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ represents the actual value from the prior period and $F_t$ is the forecast value from the previous period.

Part 1A:
If α = 0.1 and the initial forecast for October is $1.83, with the actual October value being $1.57.

Thus, the forecast for period 11 is calculated as:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Consequently, the forecast for period 11 amounts to $1.80

Part 1B:

With α = 0.1 and the forecast for November is $1.80, while the actual November value is $1.62.

Thus, the forecast for period 12 is calculated as:

$F_{12}=\alpha
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=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Thus, the forecast for period 12 amounts to $1.78

Part 2A:

Considering α = 0.3 and the initial forecast for October is $1.76, with the actual October value being $1.57.

Thus, the forecast for period 11 is calculated as:

$F_{11}=\alpha
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Therefore, the forecast for period 11 totals $1.70


Part 2B:

Given α = 0.3 with the forecast for November at $1.70, the actual November value is $1.62.

Thus, the forecast for period 12 is calculated as:

$F_{12}=\alpha
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Part 3A:

Given α = 0.5 and the initial forecast for October stands at $1.72, with the actual value for October being $1.57.

Thus, the forecast for period 11 is:

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In the case of α = 0.5 and the forecast for November of $1.65, October’s actual value is $1.62.

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=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

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The mean absolute deviation of a forecast equals the sum of absolute values of the difference between actual values and forecasted values, divided by the count of items.

Hence, taking the actual values for October, November, and December as: $1.57, $1.62, $1.75

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Part 5:

The mean absolute deviation of a forecast is acquired by summing the absolute value of the differences between actual values and forecasted values, then dividing by the number of instances.

So, with actual values for October, November, and December as: $1.57, $1.62, $1.75

using α = 0.3 leads to forecasted values of October, November, and December: $1.76, $1.70, $1.68

Therefore, the mean absolute deviation is:

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Part 6:

The mean absolute deviation of a forecast is computed as the sum of the absolute differences between actual values and forecasted values, divided by the number of observations.

Accordingly, with actual values for October, November, and December as: $1.57, $1.62, $1.75

utilizing α = 0.5 provides forecasted values for October, November, and December: $1.72, $1.65, $1.64

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