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MrRissso [65]
4 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]4 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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3 years ago
A​ student's tuition was ​$2628. A loan was obtained for 5​/4 of the tuition. How much was the​ loan? The​ student's loan was fo
prohojiy [21]

Answer:

The loan amount of the tuition fee is $3285

Step-by-step explanation:

Given as :

The student's tuition fee amount = $2628

The loan amount for the fee = \dfrac{5}{4} of the tuition fee

Let The loan amount = $ A

So, According to question

The loan amount for the fee = \dfrac{5}{4} × tuition fee amount

A =  \dfrac{5}{4} × $2628

or, A = \frac{5\times 2628}{4}

Or, A = 5 × 657

∴  A = $3285

So, The loan amount = A = $3285

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4 0
3 years ago
Solve the equation 5x + 3 = -3x + 99 algebraically. Answer as a reduced fraction only.
abruzzese [7]

Answer:

12

Step-by-step explanation:

add 3x both sides

8x +3 = 99

subtract 3 on both sides

8x = 96

divide by 8

x = 96/8 = 12

7 0
2 years ago
Suppose that the cost of producing n pairs of shoes is given by the expression c(n)=30+5n. the average cost of producing a pair
makkiz [27]
<h3>Given</h3>

c(n)=30+5n\\a(n)=\dfrac{c(n)}{n}

<h3>Find</h3>

\dfrac{d}{dn}(a(n))\quad\text{at n=10}

<h3>Solution</h3>

\dfrac{d}{dn}(a(n))=\dfrac{d}{dn}\left(\dfrac{30+5n}{n}\right)=\dfrac{d}{dn}\left(30n^{-1}+5\right)\\\\=-30n^{-2}\\\\=-30\cdot 10^{-2}\qquad\text{at n=10}\\\\=-0.30

The derivative of the average cost function at n=10 is -0.30.

7 0
3 years ago
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