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2 years ago

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Points A, B, and C are on line AC. A horizontal line has points A, B, C. A line extends from point B up and to the left to point

D. Angle D B C is 140 degrees. Angle CBD has a measure of 140°. What is the measure of angle ABD?
a. 40°
b. 50°
c. 60°
d. 70°

2 answers:

Debora [2.8K]2 years ago

8 0

Answer:40 degrees

Step-by-step explanation:

Ksju [112]2 years ago

6 0

Answer

A. 40° on e2020

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Bartolo is running a race. He determines that if he runs at an average speed of 15 feet/second, he can finish the race in 6 seco

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Answer:

C

Step-by-step explanation:

the distance is 15•6 = 18•5 = 90 feet

so, time needed = 90/30 = 3 second

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1 year ago

WHEN GIVEN A RAW SCORE, EXPLAIN HOW TO USE THE NORMAL CURVE TO COMPARE THAT SCORE TO THE POPULATION. WHAT IS A Z-SCORE?

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<span>1. </span>When given a raw score, it must be converted into a z-score (standard score). Raw scores cannot be placed on a normal distribution curve because they do not have the same means and standard deviations, but when it is converted into a z-score, the number of standard deviations above or below the population mean can be measured. The z-scores on the center are average, the scores on the left are lower than average and the scores on the right are higher than average.

<span>2. </span>A z-score is a standard score which can be placed on a normal distribution curve. A z-score indicates the distance of the standard deviations from the mean (center of the curve).

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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>

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1 year ago

Students make 82.5 ounces of liquid soap for a craft fair. They put the soap in 5.5-ounce bottles and sell each bottle for $4.

Dovator [93]

Answer:

$63

Step-by-step explanation:

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1 year ago

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A movie theater sells popcorn in bags of different sizes. The table shows the volume of popcorn and the price of the bag.

scoundrel [369]

Answer:

<em>The price for a 60-ounce bag of popcorn would be $16</em>

Step-by-step explanation:

<u>Function Modeling</u>

The behavior of some parameters that depend on a set of variables can be modeled in several ways, like linear, quadratic, exponential, logarithmic, among many others.

The selection of the model is often a complex decision that involves statistics and data analysis.

The question provides us with four points where the volume of popcorn bags and the price in dollars. The easiest function that can be used is the line.

The equation of a line of the volume V and the price p can be found with the expression

$\displaystyle p-p_1=\frac{p_2-p_1}{V_2-V_1}(V-V_1)$

We'll use the first two values (6,10) (8,20)

$\displaystyle p-6=\frac{8-6}{20-10}(V-10)$

Simplifying and rearranging, we get the model

$\displaystyle p(V)=\frac{1}{5}V+4$

To test the accuracy of the model, we compute the values of p for V=35 and for V=48

$\displaystyle p(35)=\frac{1}{5}(35)+4=11$

$\displaystyle p(48)=\frac{1}{5}(48)+4=13.6$

Since the computed values are equal to those of the table, the model is accurate. We can now predict the price for V=60

$\displaystyle p(60)=\frac{1}{5}(60)+4=16$

The price for a 60-ounce bag of popcorn would be $16

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1 year ago