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Rom4ik [11]
3 years ago
11

During one month there were 7 days of precipitation . what if there had only been 3 days of precipitation that month? How would

that change the measures of center?
Mathematics
2 answers:
zmey [24]3 years ago
8 0

Answer:

The Mean will decrease while the Median and the Mode will remain with the same amount.

Step-by-step explanation:

For Central Measures we have Median, Mode and  Mean. They try to describe the whole distribution by one value, in the center of it.

For this question, let's pick some examples from the Old Faithful, WY Station to the National Oceanic and Atmospheric Administration from 2015, January

1st Case Scenario:

For 7 days of precipitation, (Snow, ice pellets, etc. rain) in inches:

19 19 19 24 23 23 23 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Mean \mu=5\\

Median=0 (the sum of the 15th+16th position over 2)

Mode=0 (Most common observations)

2nd Case Scenario

For 3 days of precipitation, (Snow, ice pellets, etc. rain) in inches:

0 0 0 24 0 0 23 0 19 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Mean = 2.2

Median =0

Mode=0

docker41 [41]3 years ago
6 0
Answer: The measures of center would most likely get smaller.

I assume that we are looking for the average rainfall during a month. If we had 7 days, there would be more rain meaning higher measures of center.

On the other hand, with only 3 days, they would most likely be lower. Unless, we had a lot of rain on those 3 days.
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hichkok12 [17]

The equation that has the same solution as  2.3p – 10.1 = 6.5p – 4 – 0.01p are as follows;

  • 2.3p – 10.1 = 6.49p – 4
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<h3>How to rewrite an equation? </h3>

2.3p – 10.1 = 6.5p – 4 – 0.01p

The equation that has the same solution as above can be found as follows;

2.3p – 10.1 = 6.5p – 4 – 0.01p

combine like terms

2.3p – 10.1 = 6.5p – 0.01p - 4

2.3p – 10.1 = 6.49p – 4

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learn more on equation here: brainly.com/question/26005958

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