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2 years ago

If a and b are two integers then closure property states then

Mathematics

1 answer:

Margarita [4]2 years ago

4 0

Your question has been heard loud and clear.

If a and b are two integers , then closure property states that the product of the two integers a and b , which is ab , is also an integer.

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Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

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

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

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

Therefore, the foreast for period 11 is $1.80

Part 1B:

Given α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78

Part 2A:

Given α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68


Part 3A:

Given α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107


Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097

5 0

2 years ago

Which multiplication equation can be used to explain the solution to 15 divided by 1/3

Anastaziya [24]

Answer:Multiply by the reciprocal, also sometimes referred to as "Keep, Change, Flip." Here is how it works. You rewrite the division question as a multiplication question by flipping the second fraction over. Next, keep the first number, change the division to multiplication and then flip the second fraction over.

Step-by-step explanation:

4 0

2 years ago

Jeffrey thinks that the largest place value that 8.21 and 19.5 have in common is the tens place Austin thinks that the largest p

Crazy boy [7]

Answer:

Austin is correct.

Step-by-step explanation:

19.5 has a tens, ones, and tenths place. 8.21 has a ones, tenths, and hundredths place.

8 0

2 years ago

Read 2 more answers

Uma urna contém 10 bolas identificadas pelas letras A, B, ..., J. Uma bola é extraída ao acaso da urna e sua letra é observada.

arlik [135]

Answer:

a) Probability that a letter of the drawn ball is vowel = (3/10) = 0.30

b) Probability that a letter of the drawn ball is consonant = (7/10) = 0.70

a) Probabilidade de uma letra da bola sacada ser vogal = (3/10) = 0.30

b) Probabilidade de uma letra da bola sacada ser consoante = (7/10) = 0,70

Step-by-step explanation:

English Translation

A ballot box contains 10 balls identified by the letters A, B, ..., J. A ball is drawn at random from the ballot box and its letter is observed. (Make the sample space and all events explicit). What is the probability that a letter of the drawn ball is: a) Vowel b) Consonant

Solution

The 10 balls are identified by A, B, C, D, E, F, G, H, I and J

Note that the probability of an event is given as the number of elements in that event divided by the number of elements in the sample space.

a) Probability of drawing a ball that has a vowel letter = P(v)

P(v) = n(v) ÷ n(S)

n(v) = Number of balls with vowel letters = 3 (that is, A, E and I)

n(S) = Total number of balls = 10

P(v) = (3/10) = 0.30

b) Probability of drawing a ball that has a consonant letter = P(c)

P(c) = n(c) ÷ n(S)

n(c) = Number of balls with consonant letters = 7 (that is, B, C, D, F, G, H and J)

n(S) = Total number of balls = 10

P(c) = (7/10) = 0.70

In Portugese/Em português

As 10 bolas são identificadas por A, B, C, D, E, F, G, H, I e J.

Observe que a probabilidade de um evento é fornecida como o número de elementos nesse evento dividido pelo número de elementos no espaço de amostra.

a) Probabilidade de desenhar uma bola com uma letra de vogal = P (v)

P (v) = n (v) ÷ n (S)

n (v) = Número de bolas com letras de vogal = 3 (ou seja, A, E e I)

n (S) = Número total de bolas = 10

P (v) = (3/10) = 0,30

b) Probabilidade de desenhar uma bola com uma letra consoante = P (c)

P (c) = n (c) ÷ n (S)

n (c) = Número de bolas com letras consoantes = 7 (ou seja, B, C, D, F, G, H e J)

n (S) = Número total de bolas = 10

P (c) = (7/10) = 0,70

Hope this Helps!!!!

Espero que isto ajude!!!!

3 0

2 years ago

Which expression represents the total perimeter of her sandwich, and if x = 1.2, what is the approximate length of the crust?

soldi70 [24.7K]

The answer to this question is choice option letter C

7 0

2 years ago

Read 2 more answers

If a and b are two integers then closure property states then​

If a and b are two integers then closure property states then