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Crank
3 years ago
8

A group of adults and students went on a class trip to Washington, DC. The number of male students was 1 more than 7 times the n

umber of adults. The number of female students was half the number of male students. If the total number of people who went on the trip is 82, find the numbers of male students and female students.
Mathematics
1 answer:
Helga [31]3 years ago
8 0
In equation, let x be the number of male students
a be the number of adults
y be the number of female students.
x= 7a+1
a= x/7 -1
y= x/2 or (7a + 1)/ 2

a + b = 82, let b be the number of students.
a + (x + y) = 82
a + [7a+1 + (7a+1)/2] = 82
a + [{2(7a+1) + 7a+1} / 2] = 82
a + [(14a +2 + 7a +1) / 2] = 82
a + [(21a + 3) / 2] = 82
(2a+ 21a + 3) / 2 = 82
(23a + 3) / 2 = 82
23a + 3 = 164
23a = 164 -3 
23a = 161
a = 7

x = 7(7) +1, 49+1 = 50 male students
y=x/2, 50/2, 25 female students

50(male students) + 25(female students)  + 7 (adults) = 82





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a) r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1  

We have a perfect linear relationship between the two variables

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And we can find the intercept using this:  

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So the line would be given by:  

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c) For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

Step-by-step explanation:

We have the following data:

Number of hours spent practicing (x) 0 0.5 1 1.5 2 2.5 3 3.5 4

Score in the game (y) 5 8 11 14 17 20 23 26 29

Part a

The correlation coefficient is given:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=9 \sum x = 18, \sum y = 153, \sum xy = 396, \sum x^2 =51, \sum y^2 =3141  

r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1  

We have a perfect linear relationship between the two variables

Part b

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}  

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}=51-\frac{18^2}{9}=15  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=396-\frac{18*153}{9}=90  

And the slope would be:  

m=\frac{90}{15}=6  

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

\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2  

\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17  

And we can find the intercept using this:  

b=\bar y -m \bar x=17-(6*2)=5  

So the line would be given by:  

y=6 x +5  

Part c

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And the intercept 5 represent the minimum score expected for any game

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