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ser-zykov [4K]
3 years ago
10

Can some one help me find the minimum, third quartile, first quartile, median and the max?

Mathematics
1 answer:
Kipish [7]3 years ago
5 0
I'll walk you through the first one, and then you should be able to do the rest. 

The first step is to right out your numbers from least to greatest

Problem #1: 2,4,5,6,8,10,13,17,19,20

To find the MEDIAN, you're simply crossing out a number from each end until you meet in the middle. In this case, you have an even number of data, which means once you get to the middle, you're have to find the average.

In this problem, your two middle numbers are 8 & 10. Since only one number can be the median, you add them together, and divide by 2:

8+10=18  18/2=9 < this is your middle, or MEDIAN 

Next, your first and third quartiles--  they're the median of the lower half and data, and upper half. 

For the lower quartile, find the mean of 2,4,5,6, and 8. again, cross out one from each side until you get to the middle.  FIRST QUARTILE = 5

Do the same process for the upper half of data (10,13,17,19 & 20).
THIRD QUARTILE = 17

The MIN is the lowest number of data = 2
The MAX is the highest number of data = 20

Best of luck!
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Answer:

(5,1) is not a solution of the above system of equations.

Step-by-step explanation:

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So, solving equation (1) we get y = 1 and putting y = 1 in the equation (2) we get 7x = 7 - 6(1) = 1

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