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If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean
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22 Oct 2018, 03:58
22 Oct 2018, 03:58
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If \(x > 0\) and \(2x^2 + 6x = 8\), then the average (arithmetic mean) of x + 2, 2x – 1, and x + 4 is equal to which of the following?
A. –2
B. 3
C. 3.5
D. 5
E. 7
A. –2
B. 3
C. 3.5
D. 5
E. 7
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Posts: 6218
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Re: If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean
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22 Oct 2018, 09:06
22 Oct 2018, 09:06
1
Carcass wrote:
If \(x > 0\) and \(2x^2 + 6x = 8\), then the average (arithmetic mean) of x + 2, 2x – 1, and x + 4 is equal to which of the following?
A. –2
B. 3
C. 3.5
D. 5
E. 7
A. –2
B. 3
C. 3.5
D. 5
E. 7
Given: 2x² + 6x = 8
Divide both sides by 2 to get: x² + 3x = 4
Subtract 4 from both sides to get: x² + 3x - 4 = 0
Factor: (x + 4)(x - 1) = 0
Solved to get: EITHER x = -4 OR x = 1
Since we're told that x > 0, we can be certain that x = 1
QUESTION: What is the average (arithmetic mean) of x + 2, 2x – 1, and x + 4?
If x = 1, then:
x + 2 = 1 + 2 = 3
2x - 1 = 2(1) - 1 = 1
x + 4 = 1 + 4 = 5
Average of 3, 1 and 5 = (3 + 1 + 5)/3
= 9/3
= 3
Answer: B
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If x > 0 and 2x2 + 6x = 8, then the average (arithmetic mean
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08 Jul 2021, 08:50
08 Jul 2021, 08:50
1
Theory: Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\)
We need to find the average (arithmetic mean) of x + 2, 2x – 1, and x + 4
Total Number of Values= 3
Sum of All the Values= x+2 + 2x-1 + x + 4 = x + 2x + x + 2 -1 +4 = 4x + 5
Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\) = \(\frac{4x+5}{3}\)
Let's solve for x now
\(x > 0\) and \(2x^2 + 6x = 8\)
=> \(2x^2 + 6x - 8 = 0\)
Divide both sides by 2 we get
\(x^2 + 3x - 4 = 0\)
=> \(x^2 - x + 4x - 4 = 0\)
=> \(x(x - 1) + 4(x - 1) = 0\)
=> \((x - 1) (x+ 4) = 0\)
=> x = 1, -4
Since, x > 0
=> x = 1
=> Average = \(\frac{4x+5}{3}\) = \(\frac{4*1+5}{3}\) = \(\frac{9}{3}\) = 3
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
We need to find the average (arithmetic mean) of x + 2, 2x – 1, and x + 4
Total Number of Values= 3
Sum of All the Values= x+2 + 2x-1 + x + 4 = x + 2x + x + 2 -1 +4 = 4x + 5
Average = \(\frac{Sum Of All The Values}{Total Number Of Values}\) = \(\frac{4x+5}{3}\)
Let's solve for x now
\(x > 0\) and \(2x^2 + 6x = 8\)
=> \(2x^2 + 6x - 8 = 0\)
Divide both sides by 2 we get
\(x^2 + 3x - 4 = 0\)
=> \(x^2 - x + 4x - 4 = 0\)
=> \(x(x - 1) + 4(x - 1) = 0\)
=> \((x - 1) (x+ 4) = 0\)
=> x = 1, -4
Since, x > 0
=> x = 1
=> Average = \(\frac{4x+5}{3}\) = \(\frac{4*1+5}{3}\) = \(\frac{9}{3}\) = 3
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
