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bulgar [2K]
3 years ago
6

For isosceles trapezoid ABCD, find the length of the median and m∠A

Mathematics
2 answers:
TiliK225 [7]3 years ago
5 0

Answer:

10

68

Step-by-step explanation:

The median joins the midpoints of AD and CB. It's formula is 1/2 * (5 + 15) = 10 for its length.

If the trapezoid is isosceles, then <A is the same size as <B

frozen [14]3 years ago
3 0

Answer:

The length of the median is 10, and the measure of the angle ∡A is 68°.

Step-by-step explanation:

Let us start by the median. Recall that the median of a trapezoid is the segment that joins the midpoints of the non-parallel sides. In this case, the median will join the midpoints of the segments DA and CB. The length of the median of a trapezoid can be easily calculated by the formula

m=\frac{B+b}{2}

where m stands for the median, B for the larger of the parallel sides, and b for the shorter one. In this particular case B=AB and b=DC. Thus,

m=\frac{AB+DC}{2} = \frac{15+5}{2} = 10.

Finally, recall that one of the main properties of isosceles trapezoid is that the angles adjacent to the parallel sides are equal. Then, as ∡B=68°, we conclude that ∡A is 68°.

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Factor 9y2 - 12y + 4.<br><br> (3 y - 2)(3 y + 2)<br> (3 y - 4)( y + 1)<br> (3 y - 2) 2
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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:

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Thus, the forecast for period 11 is given by:

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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

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</span></span>
<span><span>Part 6:

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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

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&#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>
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