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lapo4ka [179]
3 years ago
9

Charlie used a regression calculator to generate the equation f(x) = –0.15x + 20.1 for the ordered pairs (2, 15), (4, 21), (6, 2

6), (8, 20), and (10, 14). Is a linear representation the best way to represent the data? If it is, explain why. If not, explain why and suggest a better alternative
Mathematics
2 answers:
aivan3 [116]3 years ago
8 0
Answer: The r<span>-value for the linear function related to the ordered pairs is very close to zero, so it is not a good representation of the data. A quadratic model would better represent the data because there is a turning point within the data set. The data increases then decreases, which is what the graph of a quadratic does. </span>
Crazy boy [7]3 years ago
4 0

Answer:

The r-value for the linear function related to the ordered pairs is very close to zero, so it is not a good representation of the data. A quadratic model would better represent the data because there is a turning point within the data set. The data increases then decreases, which is what the graph of a quadratic does.

Step-by-step explanation:

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Answer:

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And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X         5       6         7       8        9        10

P(X)   0.05   0.15  0.33  0.28   0.12   0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:  

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :  

E(X^2)=\sum_{i=1}^n X^2_i P(X_i)  

And using the formula we got:  

E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46  

Then we can find the variance with the following formula:  

Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096  

And then the standard deviation would be given by:  

Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229  

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