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2 years ago

What is msu? 49° 77° 98° 161°

Mathematics

2 answers:

Trava [24]2 years ago

6 0

Answer:

Step-by-step explanation:

I just took the test !!

givi [52]2 years ago

4 0

Answer:

77°

Step-by-step explanation:

$\\ T = \frac{1}{2} (RU - SU)\\ \\ 21 =\frac{1}{2}(119 - SU)\\ \\ 42 = 119 - SU\\ SU=77$

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ΔRST is dilated about the origin, O, to create ΔR'S'T'. Point R is located at (3, 9), and point R' is located at (1.2, 3.6). Whi

motikmotik

Answer 
scale factor=0.4

Explanation 
The scale factor for any transformation is the ration of the image distance to the object distance.
That is the ratio of distance from the origin to the image to the distance from the origin to the object.
In the case above, OR’is the image distance and OR is the object distance.
OR’ = √(〖1.2〗^2+〖3.6〗^2 )=√14.4=3.794733192
OR = √(3^2+9^2 )=√90=9.486832981

The scale factor = 3.794733192/9.486832981
=0.4

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2 years ago

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lbvjy [14]

Equivalent equations are equations that have exactly the same solution. To find equivalent equations, it is best to solve them individually to find the solutions. The ones with the same solution in the end are considered equivalent equations.

Hope this helps!

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2 years ago

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The yearbook staff enlarges a picture with a length of 5 inches and a width of 7

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I believe yes, the area of the picture is 78.75 square inches. hope this helps you

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In 24 hours a baby drank 6 bottles of formula. Each bottle holds 8 fluid ounces. How many cups of formula did the baby drink

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Answer:

6 cups

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2 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 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