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kompoz [17]
3 years ago
9

1. What does this quote mean to you?

Mathematics
2 answers:
Effectus [21]3 years ago
5 0

This quote is telling us that you can't exist unless you are doing something worth being remembered for. Through out our lives we all listen to what the world tells us, but not many of us actually do what we want to. We live our lives for someone else, but I plan to change that. Instead of letting the public decide how I should live my life I will live for me. I will work harder in school, become smarter than my parents and grow as a person. I want to be remembered for stepping out of my comfort zone and being different. This is the year that I exist, and show people that I exist for a purpose.

Step-by-step explanation:

Went really deep with this one.

insens350 [35]3 years ago
4 0

Answer:

My opinion is that you can exist any many things as you navigate through life. This quote is very wrong because you can exist in doing things for other people and not just yourself! You can not be selfish your entire life! I think that you should be for other people! I think that doing stuff for yourself is okay to a certain extent!

Step-by-step explanation:

THIS IS SOOO WRONG( idk tho)

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If there os 150 capacity and 36 os used what os the remaining percentage
Veseljchak [2.6K]
150 - 36 = 114 os remaining;
114/150 = 0,76 (as a decimal) = 76% (as a percentage);
6 0
3 years ago
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 <span>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:

</span>Given <span>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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.1(1.62)+(1-0.1)(1.80) \\  \\ &#10;=0.162+0.9(1.80)=0.162+1.62 \\  \\ =1.782

Therefore, the foreast for period 12 is $1.78</span>



Part 2A:

Given <span>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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.3(1.57)+(1-0.3)(1.76) \\  \\ &#10;=0.471+0.7(1.76)=0.471+1.232 \\  \\ =1.703

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.3(1.62)+(1-0.3)(1.70) \\  \\ &#10;=0.486+0.7(1.70)=0.486+1.19 \\  \\ =1.676

Therefore, the foreast for period 12 is $1.68



</span></span>
<span>Part 3A:

Given <span>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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.5(1.57)+(1-0.5)(1.72) \\  \\ &#10;=0.785+0.5(1.72)=0.785+0.86 \\  \\ =1.645

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.5(1.62)+(1-0.5)(1.65) \\  \\ &#10;=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 </span></span></span><span>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 </span><span>using exponential smoothing where α ​= 0.1 of October, November and December is given by: 0.157



</span><span><span>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 </span><span>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:

&#10; \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = &#10;\frac{|-0.17|+|-0.08|+|-0.07|}{3}  \\  \\ = \frac{0.17+0.08+0.07}{3} = &#10;\frac{0.32}{3} \approx0.107

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



</span></span>
<span><span>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 </span><span>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:

&#10; \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = &#10;\frac{|-0.15|+|-0.03|+|0.11|}{3}  \\  \\ = \frac{0.15+0.03+0.11}{3} = &#10;\frac{29}{3} \approx0.097

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α ​= 0.5 of October, November and December is given by: 0.097</span></span>
5 0
3 years ago
Write an algebraic expression for the given quantity. Let x represent the unknown value.
Lelechka [254]

Answer:

Step-by-step explanation:

There is no equal sign.

1/2 x - 5*x

is all there is.

7 0
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2. JK -1-2/4+3
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Oliga [24]
B.28 Years Hope it help
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