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irina1246 [14]
3 years ago
10

Can somebody explain to me how to do this

Mathematics
1 answer:
11111nata11111 [884]3 years ago
3 0

Answer:below hopes it helps the link should help aswell

Step-by-step explanation:

https://mathsolver.microsoft.com/en/solve-problem/%60frac%7B%2010%20%7B%20q%20%20%7D%5E%7B%205%20%20%7D%20%20%20%7B%20w%20%20%7D%5E%7B%207%20%20%7D%20%20%20%20%7D%7B%202%20%7B%20w%20%20%7D%5E%7B%203%20%20%7D%20%20%20%20%7D%20%20%20%60frac%7B%204%20%7B%20%60left(%20%7B%20q%20%20%7D%5E%7B%206%20%20%7D%20%20%20%60right)%20%7D%5E%7B%202%20%20%7D%20%20%20%20%7D%7B%20%20%7B%20w%20%20%7D%5E%7B%20-5%20%20%7D%20%20%20%20%7D

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A conjunction is a statement formed by combining two statements with the word ____. and together with or
Verizon [17]

Answer:

and

Step-by-step explanation:

A conjunction is formed when two statements join with the connector <u>and.</u>

The symbol used for a conjunction is:

∧

to mean AND.

For example, when you have two statements :

p : It will rain today.

q : I will fetch water.

You can connect the statements to form a compound statement :

It will rain today and i will fetch water. ( p ∧ q ).

3 0
3 years ago
What’s the answer? ANSWER THIS NOW PLEASE
levacccp [35]

Answer:

v=4

Step-by-step explanation:

-23=3v+9-11v

-23=9-8v

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5 0
3 years ago
Read 2 more answers
Find the vertex of the given quadratic function.
ki77a [65]

B

given f(x) in factored form, equate to zero for x-intercepts

(x - 8)(x - 4) = 0, hence

x = 4, x = 8 ← x- intercepts

The vertex lies on the axis of symmetry which is situated at the midpoint of the x- intercepts

x- coordinate of vertex = \frac{4+8}{2} = 6

f(6) = (6 - 8 )(6 - 4 ) = -2 × 2 = - 4 ← y-coordinate

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8 0
3 years ago
Is 11/12 greater or less than 2/3?
Sauron [17]
False beacause 2/3 is bigger than 11/12. Think about it like pizza 2/3 pieces is better cuz there bigger & 11/12 are smaller pieces.
8 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}&#10;{|c|c|c|c|}&#10;Month&Price per Chip&Month&Price per Chip\\[1ex]&#10;January&\$1.90&July&\$1.80\\&#10;February&\$1.61&August&\$1.83\\&#10;March&\$1.60&September&\$1.60\\&#10;April&\$1.85&October&\$1.57\\&#10;May&\$1.90&November&\$1.62\\&#10;June&\$1.95&December&\$1.75&#10;\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
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