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vovikov84 [41]
3 years ago
15

Melanie uses the ordered pairs (2010, 48) and (2013, 59) to find her equation. Tracy defines x as the number of years since 2010

and uses the ordered pairs (0, 48) and (3, 59) to find her equation. How will the two girls’ equations compare?
Mathematics
2 answers:
rewona [7]3 years ago
5 0
They will have same slopes but different Y intercepts.
Naily [24]3 years ago
5 0

Answer:  The solpes of the two equations are same but y-intercepts are different.

Step-by-step explanation:  Given that Melanie uses the ordered pairs (2010, 48) and (2013, 59) to find her equation.

Tracy defines x as the number of years since 2010 and uses the ordered pairs (0, 48) and (3, 59) to find her equation.

We are to compare the equations of Melanie and Tracy.

Since both the girls used two points to find their equations, so the graph of the equations must be straight lines.

<u>Melanie's Equation:</u>  Since two points in Melanie's equation are (2010, 48) and (2013, 59), so the slope of the line will be

m_1=\dfrac{59-48}{2013-2010}=\dfrac{11}{3}\\\\\\\Rightarrow m_1=\dfrac{11}{3}.

Therefore, Melanie's equation is

y-48=m_1(x-2010)\\\\\Rightarrow y-48=\dfrac{11}{3}(x-2010)\\\\\Rightarrow y=\dfrac{11}{3}x-11\times 670+48\\\\\Rightarrow y=\dfrac{11}{3}x-7322.

<u>Tracy's Equation:</u>  Since two points in Tracy's equation are (0, 48) and (3, 59), so the slope of the line will be

m_2=\dfrac{59-48}{3-0}=\dfrac{11}{3}\\\\\\\Rightarrow m_2=\dfrac{11}{3}.

Therefore, Tracy's equation is

y-48=m_2(x-0)\\\\\Rightarrow y-48=\dfrac{11}{3}x\\\\\Rightarrow y=\dfrac{11}{3}x+48.

Thus, we can see that

m_1=m_2~~~\textup{and}~~~-7322\neq 48,

so, the slopes of two equations are same but y-intercepts are different.

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Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

4 0
3 years ago
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