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Serjik [45]
3 years ago
15

ANSWER ASAP

Mathematics
1 answer:
lukranit [14]3 years ago
5 0

Use a graphing calculator in the future.

(2.286, −4.857)

A.

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The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the
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D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

  • Now if we consider the defect in sixth car to be 3 then our data will look like:    3,4,6,7,9,10

So the mean of above data would be \frac{3+4+6+7+9+10}{6} = 6.5

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{6+7}{2} = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

  • Now if we consider the defect in sixth car to be 12 then our data will look like:    4,6,7,9,10,12

So the mean of above data would be \frac{4+6+7+9+10+12}{6} = 8

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{7+9}{2} = 8

Hence mean and median number of defects are same if the sixth car has 12 defects.

  • But Now if we consider the defect in sixth car to be 7 then our data will look like:    4,6,7,7,9,10        

So the mean of above data would be \frac{4+6+7+7+9+10}{6} = 7.167    

and Median of the above data would be =  \frac{3^{rd} obs + 4^{th}  obs}{2} = 7

   Hence mean and median number of defects are not same if the sixth car has 7 defects.      

Therefore option D is correct.                                                                                  

8 0
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