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Fynjy0 [20]
3 years ago
15

Killian went to the movies with his mother, 2 brothers, and 2 sisters. The theater had choices of 2 different types of soda and

3 different types of candy. If each person ordered a soda and box of candy, is it possible for each person in Killian's family to order a different combination? Be sure to show your work and explain your answer for full credit!!
Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0

yes it is possible for them all to have a different combination

his brother could have soda A and candy A

while the other brother had soda B and candy B

then his sister could have soda C and candy B

then his other sister could have soda C and candy A

while finally killian could have soda B and candy A

 

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Step-by-step explanation:

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

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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}

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\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

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