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lorasvet [3.4K]
3 years ago
15

Help fast timed hurry please ​

Mathematics
1 answer:
NARA [144]3 years ago
6 0

Answer:

b: 4

Step-by-step explanation:

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

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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

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The MAX is the highest number of data = 20

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5 0
3 years ago
Divide<br><br><br> 1520÷83 =_ R_
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1520÷83= 18R26

First, you have to divide it like whole numbers. Then you have 3 options:

1. Stop and turn the fraction into a remainder (R15)
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LiRa [457]

Answer is in the screenshot. Hope this helped!

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3 years ago
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