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Yuki888 [10]
3 years ago
11

100 points if correct In this task, you will play the role of a restaurant owner. You will collect data and analyze different as

pects of your business.
As you complete the task, keep these questions in mind: Can you predict sales based on factors such as the weather and the number of customers? Can you estimate the amount you will earn in tips on a given day based on the number of customers?
Directions:
Complete each of the following tasks, reading the directions carefully as you go. You may use a piece of graph paper for this task.
Round all decimals to the nearest hundredths place or to the nearest cent for this task.
You will be graded on the work you show, or your solution process, in addition to your answers. Make sure to show all of your work and to answer each question as you complete the task. You will be given partial credit based on the work you show and the completeness and accuracy of your explanations.
Your teacher will give you further directions on submitting your work. You may be asked to upload the document, e-mail it to your teacher, or hand in a hard copy.
Now let’s get started!


Step 1: Constructing and analyzing scatterplots


You keep track of the daily hot chocolate sales and the outside temperature each day. The data you gathered is shown in the data table below.

Hot Chocolate Sales and Outside Temperatures
Sales ($) $100 $213 $830 $679 $209 $189 $1,110 $456 $422 $235 $199
Temperature (°F) 92° 88° 54° 62° 85° 16° 52° 65° 68° 89° 91°

a) Make a scatterplot of the data above. (2 points)

b) Do you notice clusters or outliers in the data? Explain your reasoning. (2 points)








c) How would you describe the correlation in the data? Explain your reasoning. (2 points)








d) What are the independent and dependent variables? (2 points)








Step 2: Evaluating trends of data

Because you want to prepare and serve the healthiest food possible, you monitor the fat and calorie content of items on your menu. Some of the menu items are included in the graph below.



a) Your business partner describes this as a high positive correlation. Is your partner correct? Why or why not? (2 points)






b) Using the drawing tools, draw a trend line (line of best fit) on the graph above. (2 points)




c) Judge the closeness of your trend line to the data points. Do you notice a relationship between the data points? (2 points)






d) Is the trend line linear? If so, write a linear equation that represents the trend line. Show your work. (3 points)










Step 3: Making predictions using data

You and your business partner track the number of customers served and the amount of tips collected per day. The data you gathered is displayed in the chart below.

Servers’ Collected Tips
Customers 54 46 34 67 52 22 49 64 55 80 38 42
Tips ($) $92 $80 $76 $121 $109 $43 $87 $114 $99 $174 $88 $91

a) Create a scatterplot displaying the data in the table. Be sure to include a linear trend line. (2 points)







b) Find the equation of the trend line (line of best fit). Show your work. (2 points)



















c) Predict the amount of tips that would be collected if 100 customers were served at the restaurant on a given day. Explain your reasoning. (2 points)


d) Explain how to use the regression calculator to make a reasonable prediction given a data table. (2 points)
Mathematics
1 answer:
jek_recluse [69]3 years ago
7 0

Answer:

Step-by-step explanation:

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As the ladder is pulled away from the wall, the area and the height with the

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The correct response are;

  • (a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

  • (b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

  • (c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, \displaystyle \frac{dx}{dt} = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}

25² = x² + y²

y = √(25² - x²)

\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}

Which gives;

\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times \frac{dx}{dt} =  \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2

\displaystyle \frac{dy}{dt} =  \mathbf{ \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2}

When x = 15, we get;

\displaystyle \frac{dy}{dt} =   \frac{15 \times \sqrt{625-15^2}  }{15^2- 625}\times2 = \mathbf{-1.5}

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

\displaystyle \frac{dy}{dt} =   \frac{20 \times \sqrt{625-20^2}  }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6

The velocity of the top of the ladder = \underline{-2.\overline{6} \ m/s \ downwards}

When x = 24, we get;

\displaystyle \frac{dy}{dt} =   \frac{24 \times \sqrt{625-24^2}  }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}}  \approx -6.86

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) \displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}

Therefore;

\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}

\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}

\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}

Therefore;

\displaystyle \frac{dA}{dt} =  \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2

When the ladder is 24 feet from the wall, we have;

x = 24

\displaystyle \frac{dA}{dt} =  \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}

The rate the area formed by the ladder is changing, \displaystyle \frac{dA}{dt} ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

\displaystyle sin(\theta) = \frac{x}{25}

\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}

\displaystyle \frac{d \theta}{dt}  = \frac{d \theta}{dx} \times \frac{dx}{dt}

\displaystyle\frac{d \theta}{dx}  = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}

Which gives;

\displaystyle \frac{d \theta}{dt}  =  -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}

When x = 24 feet, we have;

\displaystyle \frac{d \theta}{dt} =  -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is \displaystyle \frac{d \theta}{dt} ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0
2 years ago
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