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The average (arithmetic mean) of the values in d
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Updated on: 02 May 2023, 09:58
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The average (arithmetic mean) of the values in data set R is m. Data set S consists of the values in data set R and the value k. and the average of the values in data set S is m + a. where a > 0. Data set T consists of the values in data set S and the value k.
Quantity A
Quantity B
The average of the values in data set T
m + 2a
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given.
Originally posted by beretta on 01 May 2023, 23:59.
Last edited by beretta on 02 May 2023, 09:58, edited 1 time in total.
Re: The average (arithmetic mean) of the values in d
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02 May 2023, 09:41
Please check and correct this part of the question
beretta wrote:
The average (arithmetic mean) of the values in data set R is in. Data set S consists of the values in data set R and the value k. and the average of the values in data set S is m + a. where a > 0. Data set T consists of the values in data set S and the value k.
Re: The average (arithmetic mean) of the values in d
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07 May 2023, 05:45
Given that The average (arithmetic mean) of the values in data set R is m. Data set S consists of the values in data set R and the value k. and the average of the values in data set S is m + a. where a > 0. Data set T consists of the values in data set S and the value k.
Let data set R has x values Given that average of the values in data set R is m => Sum of values in data set R = Average * x = m*x = mx
=> Data Set S = {R ,k} with mean m + a Sum of values in data set S = Mean * (x+1) [ as data set R had x values ] = (m+a) * (x+1) = Sum of values in data set R + k = mx + k => mx + ax + m + a = mx + k => k = ax + m + a
=> Data Set T = {S ,k} Sum of values in data set T = Sum of values in data set S + k = mx + k + k = mx + 2*(ax + m + a) = mx + 2ax + 2m + 2a
The average of the values in data set T = \(\frac{Sum }{ x + 2}\) = \(\frac{mx + 2ax + 2m + 2a }{ x + 2}\) = \(\frac{mx + 2ax + 2m + 2a + 2a - 2a }{ x + 2}\) = \(\frac{mx + 2ax + 2m + 4a - 2a }{ x + 2}\) = \(\frac{x*(m + 2a) + 2*(m + 2a) - 2a }{ x + 2}\) = \(\frac{(x + 2)*(m + 2a)}{ x + 2}\) - \(\frac{2a }{ x+2}\) = m + 2a - \(\frac{2a }{ x+2}\) Now a and x are positive => - \(\frac{2a }{ x+2}\) will be negative => m + 2a - \(\frac{2a }{ x+2}\) < m + 2a