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MrRissso [65]
3 years ago
10

This week, we are covering relationships that can be approximated by linear equations. For instance, y = 453x + 3768 represents

the increase in cost of tuition and fees at public four-year universities. Now think about how this might apply in your chosen field and answer the following question: What is an example of a relationship (in your chosen field) that can be represented by a linear equation? Show the equation and explain the relationship. Please provide a reference for your example (website, book, etc.)
Mathematics
1 answer:
lana [24]3 years ago
8 0

Answer:

See explanation below.

Step-by-step explanation:

We assume that the data is given by :

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

Where X represent the cost for scholarships in thousands of dollars and y represent the cost of life for an academic semester (The data comes from the web)

We can find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720

\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324

\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200

\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540

\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900

And the slope would be:

m=-\frac{1900}{6000}=-0.317

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60

\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27

And we can find the intercept using this:

b=\bar y -m \bar x=27-(-0.317*60)=46.02

So the line would be given by:

y=-0.317 x +46.02

We have an inverse linear relationship since the slope is negative between the variables of interest.

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Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
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Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

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Step-by-step explanation:

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