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irinina [24]
2 years ago
5

Hamburgers cost $2.50 and cheeseburgers cost $3.50 at a snack bar. Ben has sold no more than $30 worth of hamburgers and cheeseb

urgers in the first hour of business. Let x represent the number of hamburgers and y represent the number of cheeseburgers. The inequality 2.50x + 3.50y < 30 represents the food sales in the first hour. If Ben has sold 4 cheeseburgers, what is the maximum value of hamburgers Ben could have sold?
Mathematics
1 answer:
Irina18 [472]2 years ago
8 0

Given 2.50x + 3.50y < 30.

Where x represent the number of hamburgers and y represent the number of cheeseburgers.

Now question is to find the maximum value of hamburgers Ben could have sold when he has sold 4 cheeseburgers.

So, first step is to plug in y=4 in the given inequality. So,

2.50x+3.50(4)<30

2.50x+14 <30

2.50x<30- 14 Subtracting 14 from each sides.

2.50x< 16

\frac{2.50x}{2.50} Dividing each sides by 2.50.

x<6.4

Now x being number of hamburgers must be an integer , so tha maximum value of x can be 6,

thus x = 6 hamburgers

So, the maximum value of hamburgers Ben could have sold is 6*2.5=$15

Hope this helps!!

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\displaystyle y=-0.927x+13.63

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<u>Simple Linear Regression </u>

It a function that represents the relationship between two or more variables in a given data set. It uses the method of the least-squares regression line which minimizes the error between the estimate function and the real data.

Let's compute the best-fit line for the data

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First, we find the sums

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\displaystyle \sum y=14+11+4+2+1=32

Then, we compute the averages values

\displaystyle \bar{x}=\frac{39}{5}=7.8

\displaystyle \bar{y}=\frac{32}{5}=6.4

We will also compute the sums of the cross-products and the sum of the squares

\displaystyle \sum xy=(1)(14)+(5)(11)+(6)(4)+(12)(2)+(15)(1)=137

\displaystyle \sum x^2=1^2+5^2+6^2+12^2+15^2=1+25+36+144+225

\displaystyle \sum x^2=431

We will compute Sxy and Sxx

\displaystyle S_{xy}=\sum xy-\frac{\sum x\ \sum y}{n}

\displaystyle S_{xy}=137-\frac{(39)(32)}{5}

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\displaystyle S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}

\displaystyle S_{xx}=431-\frac{39}{5}^2=126.8

The slope of the linear regression function is given by

\displaystyle m=\frac{S_{xy}}{S_{xx}}=\frac{-117.6}{126.8}=-0.927

The y-intercept ot the linear function is

\displaystyle b=\bar{y}-b\bar{x}=6.4-(-0.927)(7.8)

\displaystyle b=13.63

Thus the best-fit line is

\displaystyle y=-0.927x+13.63

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