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borishaifa [10]
3 years ago
14

Find the equation of the ellipse with the following properties.

Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0
The equation of an ellipse where the major axis 2a is greater than the minor 2b
 (x-h)²/a² + (y-k)²/b² =1.
When 2b>2a , then the equation becomes (y-h)²/b² + (x-k)²/a² =1
In our example a = 4 and b = 9. Moreover h=0 and k=0 because the center is on the origin, so the equation becomes:
y²/b² + x²/a² = y²/81 + x²/16 = 1
kaheart [24]3 years ago
5 0
Check the picture below

so.. hmm notice the center h,k and the major and minor axis components, thus just plug them in.

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Find the greatest common factor of 6 and 15. a. 6 b. 3 c. 4 d. 2
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Answer:

3

Step-by-step explanation:

Greatest common factor (GCF) of 6 and 15 is 3. We will now calculate the prime factors of 6 and 15, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 6 and 15.

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Pennsylvania Refining Company is studying the relationship between the pump price of gasoline and the number of gallons sold. Fo
Semenov [28]

Complete Question

Pennsylvania Refining Company is studying the relationship between the pump price of gasoline and the number of gallons sold. For a sample of 14 stations last Tuesday, the correlation was 0.65.At the 0.01 significance level Can the company conclude that the correlation is positive

Answer:

Yes the company conclude that the correlation is positive

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  14

    The correlation is  r =  0.65

     

The  null hypothesis is  H_o :  r <  0

The  alternative hypothesis is  H_1 :  r > 0

Generally the standard deviation is mathematically evaluated as

       Sr =  \sqrt{1- r}

       Sr =  \sqrt{1- 0.65}

       Sr = 0.616

The  degree of freedom for the one-tail test is

       df =  n- 2

        df =  14- 2

        df = 12

The standard error is evaluated as

        SE  =  \frac{0.616}{ \sqrt{12} }

        SE  =0.1779

The test statistics  is evaluated as

      t =  \frac{r }{SE}

       t =  \frac{0.65 }{0.1779}

        t =  3.654

The p-value of of  t is obtained from the z table, the value is  

        p-value =  P(t <  3.654) =  0.00012909

Given that p-value  <  \alpha  then we reject the null hypothesis

           Hence the company can conclude that the correlation is positive

     

6 0
3 years ago
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