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AlladinOne [14]
10 months ago
8

In a solar system, two comets pass near the sun, which is located at the origin. Comet E is modeled by quantity y plus 16 end qu

antity squared over 400 plus x squared over 144 equals 1 and Comet H is modeled by quantity y plus 13 end quantity squared over 144 minus x squared over 25 equals 1 comma where all measurements are in astronomical units.Part A: What are the vertices for the path of Comet E? Show your work. (4 points)Part B: Which comet travels closer to the sun? Give evidence to support your answer. (5 points)Part C: What key feature does the sun represent for both comets? Give evidence to support your answer. (6 points)Link to view the question file:///C:/Users/funkm/Downloads/Untitled%20document%20(2).pdf
Mathematics
1 answer:
Leya [2.2K]10 months ago
5 0

Explanation:

The equation of comet E is given below as

\frac{(y+16)^2}{400}+\frac{x^2}{144}=1

The equation of comet H is modelled below as

\frac{(y+13)^2}{144}-\frac{x^2}{25}=1

Given, that the Sun is located at the origin.

This is an equation of an ellipse. So, the path traveled by Comet e is an elliptic path.

Comparing with the Standard equation of ellipse: below,we will have

\frac{(y-k)^2}{b^2}+\frac{(x-h)^2}{a^2}

Whise center is

\begin{gathered} center=(h,k) \\ vertices=(h\pm a,0) \end{gathered}

By comparing coefficient, we will have

\begin{gathered} a^2=144 \\ a=12 \\ b^2=400 \\ b=20 \\ k=-16 \\ h=0 \\ (h,k)=(0,-16) \end{gathered}

Hence,

The vertex of comet E will be

\begin{gathered} (h\pm a,0) \\ (0+12,0),(0-12,0) \\ (12,0),(-12,0) \end{gathered}

The vertex of comet E is

(12,0),(-12,0)

Part C:

It is the foci forboth comets

\begin{gathered} cometE: \\ c=\sqrt{b^2-a^2} \\ c=\sqrt{400-144} \\ c=\sqrt{256} \\ c=16 \\  \\ For\text{ comet F:} \\ c=\sqrt{a^2-b^2} \\ c=\sqrt{144+25} \\ c=\sqrt{169} \\ c=13 \end{gathered}

Hence,

The foci will be

\begin{gathered} (16\pm16,0) \\ cometE \\ (0,0),(32,0) \\ cometF: \\ (13\pm13,0) \\ (0,0),(26,0) \end{gathered}

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Answer:

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Answer:

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If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

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t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

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t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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We can replace in formula (1) the info given like this:  

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=28-1=27  

Since is a one side lower test the p value would be:  

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