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xeze [42]
4 years ago
8

A manager is assessing the correlation between the number of employees in a plant and the number of products produced yearly. Th

e table shows the data: Number of employees (x) 0 25 50 75 100 125 150 175 200 Number of products (y) 10 160 310 460 610 760 910 1060 1210 Part A: Is there any correlation between the number of employees in the plant and the number of products produced yearly? Justify your answer. (4 points) Part B: Write a function that best fits the data. (3 points) Part C: What does the slope and y-intercept of the plot indicate?
Mathematics
1 answer:
Contact [7]4 years ago
8 0

Answer:

A)there is a strong positive correlation

B)A function that best fits the data is \widehat{y}=6x+10

C)Slope indicates the rate of change of Number of products per employee

y intercept indicates number of products when number of employees are 0

Step-by-step explanation:

Number of employees (x) 0  25  50  75  100  125  150  175  200

Number of products (y)   10 160 310 460 610 760 910 1060 1210

Part A: Is there any correlation between the number of employees in the plant and the number of products produced yearly?

Mean of x values = \frac{0+25 + 50 + 75 + 100 +  125 + 150 + 175 + 200}{9}=900

Mean of y values = \frac{10+160+ 310+ 460+ 610+ 760+ 910+ 1060+ 1210}{9} =610

\sum(x-\bar{x})^2= 37500

\sum(y - \bar{y})^2 = 1350000

N =No. of observations= 9

\sum(x - \bar{x})(y-\bar{y}) = 225000

R Calculation

r = \frac{\sum((x-\bar{x})(y - \bar{y}))}{\sqrt{\sum(x - \bar{x})^2(y-\bar{y})^2}}

r=\frac{225000}{\sqrt{(37500)(1350000)}}\\r= 1

So, there is a strong positive correlation

B: Write a function that best fits the data.

Regression Equation = ŷ = bx + a

b = \frac{225000}{37500} = 6\\a = \bar{y} -b \bar{x} = 610 - (6 \times 100) = 10\\ \widehat{y}= 6x + 10

So, a function that best fits the data is \widehat{y}=6x+10

C)What does the slope and y-intercept of the plot indicate?

Slope indicates the rate of change of Number of products per employee

y-intercept indicates the y-coordinate of a point where a line, curve, or surface intersects the y-axis.

So, y intercept indicates number of products when number of employees are 0 .

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Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

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Substituting into the total cost of the rectangular field, we get

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\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

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C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

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