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2 years ago

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Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo

r 5 units. If the firm uses exponential smoothing with an alpha value of .20, what should be the forecast for period 3? (Round answers to two decimal places.)

1 answer:

Zepler [3.9K]2 years ago

8 0

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period 2 is 9

Forecast can be given by

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

where

$F_{t+1}=Future Forecast$

$F_t=Present\ Period\ Forecast$

$D_t=Present\ Period\ Demand$

$\alpha =smoothing\ constant$

$F_{t+1}=5+0.2(7-5)$

$F_{t+1}=5.4$

Forecast for Period 3

$F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})$

$F_{t+2}=5.4+0.2\cdot (9-5.4)$

$F_{t+2}=6.12$

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Start with the solution
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2 years ago

Major arc CBD measures 300°Which is the radian measure of its corresponding central angle?

ella [17]

This question is not complete.

Complete Question

Major arc CBD measures 300°. Which is the radian measure of its corresponding central angle?

a)4pi/3 radians

b) 5pi/6 radians

c) 5pi/3 radians

d) 11pi/6 radians

Answer:

c) 5pi/3 radians = 5π/3 radians

Step-by-step explanation:

The corresponding central angle can be defined as the angle that is formed when there is a joining together of two radii at the center of the circle.

The circumference of a circle is given as

360° = 2πr

The major arc CBD is 300°

Therefore,

360° = 2πr , radius is not given hence , 360° = 2π

300° = Y (Unknown)

Cross multiply

360Y= (300× 2π)

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Y = 5π/3 radians

Hence the radian measure for the corresponding angle is = 5π/3 radians

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2 years ago

Read 2 more answers

Find the value of y for which the value of 1.7y+37 is 14 less than the value of 9.3y–25 HELP QUICK

Harrizon [31]

Answer:

y=10

Step-by-step explanation:

1.7y+37 is 14 less than the value of 9.3y–25

1.7y + 37 = 9.3y - 25 - 14
1.7y + 37 = 9.3y - 39
9.3y - 1.7y = 37 + 39
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Answer is 10

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2 years ago

2.5 kg of cabbages cost £1.20<br><br> Work out the cost of 3.75 kg

s2008m [1.1K]

Thank you for your question!

In order to get the cost of 1 kg of cabbage, divide the cost by the quantity:

1.20/2.5
0.48

Now, multiply the amount for 1 kg to 3.75, because you need 3.75 times the amount.

0.48 * 3.75
1.80

So, it would cost 1.80 pounds for 3.75 kg.
Hope this helped!

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2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

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2 years ago