The outside temperature dropped 23.4 °F from the time Taylor left for school to the time Taylor got home from school. It was 54.
1 answer:
You might be interested in
Answer: Option 4
Step-by-step explanation:
Similar triangles have congruent angles, so this means that the sines, cosines, and tangents should be the same.
Thus, the side lengths of the triangle we want to find should be multiples of .
- This eliminates options (1) and (3).
Between options 2 and 4, we know that the Pythagorean theorem is not satisfied by option (2), thus we should eliminate it.
This leaves us with option (4)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%0AMonth%26Price%20per%20Chip%26Month%26Price%20per%20Chip%5C%5C%5B1ex%5D%0AJanuary%26%5C%241.90%26July%26%5C%241.80%5C%5C%0AFebruary%26%5C%241.61%26August%26%5C%241.83%5C%5C%0AMarch%26%5C%241.60%26September%26%5C%241.60%5C%5C%0AApril%26%5C%241.85%26October%26%5C%241.57%5C%5C%0AMay%26%5C%241.90%26November%26%5C%241.62%5C%5C%0AJune%26%5C%241.95%26December%26%5C%241.75%0A%5Cend%7Btabular%7D)
The forcast for a period
is given by the formular
where
is the actual value for the preceding period and 
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:
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:
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:
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:
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:
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:
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:
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:
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:
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>
The forcast for a period
where
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:
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:
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:
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:
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:
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:
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:
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:
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:
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>