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Sveta_85 [38]
3 years ago
7

Identify the center and radius of the following circle (x-4)^2 +y^2=25

Mathematics
1 answer:
Marysya12 [62]3 years ago
7 0

Answer:

centre = (4, 0 ), radius = 5

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x - 4)² + y² = 25 ← is in standard form

with (h, k) = (4, 0) and r = \sqrt{25} = 5

Thus centre = (4, 0) and radius = 5

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2. There are 1500 candies in a candy dish. Each day half of the candy is taken out.
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Let U be the set of whole numbers from 5-10 inclusive. Set A= {5, 7, 9}, Set B = {5, 8, 10}. AU~B &lt;---------
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Suppose the universal set is U = all whole numbers from 1 to 9. If A = {1, 2, 4}, then Ac = {3, 5, 6, 7, 8, 9}. As we saw earlier with the expression Ac ...

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3 years ago
Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The following ar
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Answer:

A. E ( U ) = 21.5454  , E ( F ) = 8.39333

B. M ( U ) =  17.0 , M ( F ) =  18.0

C. E ( U' ) = 17.0  , E ( F' ) = 7.95384

D. T ( U ) = 9.091% , T ( F ) = 6.667%

Step-by-step explanation:

Solution:-

- Two sample sets ( U ) and ( F ) that define the concentration ( EU/mg ) of endotoxin found in urban and farm homes as follows:

             U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0

             F: 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 22.0 9.6 2.0 2.0 0.5

- To determine the mean of a sample E ( U ) or E ( F ) the following formula from descriptive statistics is used:

                         E ( X ) = Sum ( X_i ) / n

Where,

                         Xi : Data iteration

                         n: Sample size

Therefore,

                            E ( U ) =  \frac{Sum (U_i )}{n_u} \\\\E ( U ) =  \frac{6.0 + 5.0 + 11.0 + 33.0 + 4.0+ 5.0 +80.0+ 18.0+ 35.0+ 17.0+ 23.0 }{11} \\\\E ( U ) = 21.54545\\\\E ( F ) =  \frac{Sum (F_i )}{n_f} \\\\E ( F ) =  \frac{2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0+ 3.0+ 9.8+ 22.0+ 9.6+ 2.0+ 2.0+ 0.5 }{15} \\\\E ( F ) = 8.39333      

- To determine the sample median we need to arrange the data for both samples ( U ) and ( F ) in ascending order as follows:

             U: 4.0 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0 80.0

             F: 0.5 2.0 2.0 2.0 3.0 6.0 7.0 8.0 8.0 9.6 9.8 12.0 15.0 19.0 22.0

- Now find the mid value for both sets:

            Median term ( U ) = ( n + 1 ) / 2  

                                          = ( 11 + 1 ) / 2 = 12/2 = 6th term

            Median ( U ), 6th term = 17.0

            Median term ( F ) = ( n + 1 ) / 2  

                                          = ( 15 + 1 ) / 2 = 16/2 = 8th term

            Median ( F ), 8th term = 8.0

- We will now trim the smallest and largest observation from each set.

- In set ( U ) we see that smallest data corresponds to ( 4.0 ) while the largest data corresponds to ( 80.0 ). We will exclude these two terms and the trimmed set is defined as:

              U': 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0

- In set ( F ) we see that the smallest data corresponds to ( 0.5 ) while the largest data corresponds to ( 22.0 ). We will exclude these two terms and the trimmed set is defined as:

              F': 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 9.6 2.0 2.0

- We will again use the previous formula to calculate means of trimmed samples ( U' ) and ( F' ) as follows:

              E ( U' ) = \frac{5.0+ 5.0+ 6.0+ 11.0+ 17.0+ 18.0+ 23.0+ 33.0+ 35.0}{9} \\\\E ( U' ) = 17

              E ( F' ) = \frac{2.0 +2.0+ 2.0 +3.0+ 6.0+ 7.0+ 8.0+ 8.0+ 9.6+ 9.8+ 12.0+ 15.0+ 19.0}{13} \\\\E ( F' ) = 7.95384    

- The trimming percentage is known as the amount of data removed from the original sample from top and bottom of sample size of 11 and 15, respectively.

- We removed the smallest and largest value from each set. Hence, a single value was removed from both top and bottom of each data set. We can express the trimming percentage for each set as follows:

                  T ( U ) = \frac{1}{11} * 100 = 9.091\\\\T ( F ) =  \frac{1}{15} * 100 = 6.667%

- The trimming pecentages for each data set are 9.091% and 6.667% respectively.

7 0
3 years ago
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