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1 year ago

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A rectangle is transformed according to the rule R0, 90º. The image of the rectangle has vertices located at R'(–4, 4), S'(–4, 1

), P'(–3, 1), and Q'(–3, 4). What is the location of Q?
(–4, –3)
(–3, –4)
(3, 4)
(4, 3)

2 answers:

mars1129 [50]1 year ago

6 0

<span>(4, 3) Is your answer. </span>

3241004551 [841]1 year ago

4 0

Answer:

d

Step-by-step explanation:

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If LaTeX: m\angle ABF=8s-6m ∠ A B F = 8 s − 6 and LaTeX: m\angle ABE=2\left(s+11\right)m ∠ A B E = 2 ( s + 11 ), find LaTeX: m\a

kolbaska11 [484]

Answer:

<h2><em>2(3s-14)</em></h2>

Step-by-step explanation:

Given the angles ∠ABF=8s-6, ∠ABE = 2(s + 11), we are to find the angle ∠EBF. The following expression is true for the three angles;

∠ABF = ∠ABE + ∠EBF

Substituting the given angles into the equation to get the unknown;

8s-6 = 2(s + 11)+ ∠EBF

open the parenthesis

8s-6 = 2s + 22+ ∠EBF

∠EBF = 8s-6-2s-22

collect the like terms

∠EBF = 8s-2s-22-6

∠EBF = 6s-28

factor out the common multiple

∠EBF = 2(3s-14)

<em></em>

<em>Hence the measure of angle ∠EBF is 2(3s-14)</em>

8 0

2 years ago

The graph below shows the line of best fit for data collected on the number of cell phones and cell phone cases sold at a local

DiKsa [7]

Answer:

C

Step-by-step explanation:

The equation of a line is given by

and

slope (m) is given by:

Where (x_1,y_1) are the first set of point in the line and (x_2,y_2) is the second set of point

Let's take 2 points arbitrarily. (0,0) & (25,20)

Let's plug it and find the equation:

Now

C is the correct answer.

6 0

2 years ago

Stan's cookie recipe makes 242424 cookies and calls for exactly 384384384 sprinkles. He is wondering how many sprinkles (p)(p)le

arlik [135]

If these numbers are correct:

For every 242,424 cookies he needs 384,384,384 sprinkles. So how many sprinkles does he need for 606,060 cookies?

242,424/384,384,384 = 606,060/x

384,384,384 * 606,060 = 242,424 * x

232,959,999,767,040 = 242,424 * x

960,960,960 = x

7 0

2 years ago

Andrea bought tacos from a food truck and left a 25 % 25%25, percent tip of $ 2.00 $2.00dollar sign, 2, point, 00. What was the

Murljashka [212]

Answer:

For this case we can make the following rule of three:

25% ----------------> 2 $

100% --------------> x

From here, we clear the value of x.

We have then:

Therefore, the value of the Taco that Andrea bought, before the tip, is $ 8.

the price of Andrea's tacos, before tip is $ 8.

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1 year 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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

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

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</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:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

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 </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:

$
 \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 </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:

$
 \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 </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

1 year ago