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If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero? A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\)
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Official Solution:If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero? A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\) Rearrange the given equation: \(x^22x+15=m\). Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x) \gt 0\). This equation holds true for \(5 \lt x \lt 3\). Since \(x\) is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2. So, the probability is \(\frac{7}{21}=\frac{1}{3}\). Answer: B
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08 Nov 2014, 12:24
Bunuel wrote: Official Solution:
If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero?
A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\)
Rearrange the given equation: \(x^22x+15=m\). Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x) \gt 0\). This equation holds true for \(5 \gt x \gt 3\). Since \(x\) is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2. So, the probability is \(\frac{7}{21}=\frac{1}{3}\).
Answer: B This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3.



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09 Nov 2014, 06:09
parameswaranprasad wrote: Bunuel wrote: Official Solution:
If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero?
A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\)
Rearrange the given equation: \(x^22x+15=m\). Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x) \gt 0\). This equation holds true for \(5 \gt x \gt 3\). Since \(x\) is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2. So, the probability is \(\frac{7}{21}=\frac{1}{3}\).
Answer: B This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3. There was a typo. Should have been \(5 \lt x \lt 3\). Edited. Thank you.
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06 Jul 2015, 05:08
Bunuel wrote: parameswaranprasad wrote: Bunuel wrote: Official Solution:
If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero?
A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\)
Rearrange the given equation: \(x^22x+15=m\). Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x) \gt 0\). This equation holds true for \(5 \gt x \gt 3\). Since \(x\) is an integer then it can take the following 7 values: 4, 3, 2, 1, 0, 1, and 2. So, the probability is \(\frac{7}{21}=\frac{1}{3}\).
Answer: B This equation doesn't look right −5>x>3 . Function wouldn't hold true for values greater than 3. There was a typo. Should have been \(5 \lt x \lt 3\). Edited. Thank you. Hi Bunuel, Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because x^2 is negative in the original equation x^22x+15. *** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<5 and x>3), not IN ranges (5<x<3) . So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are 5<x<3 (I already studied the inequalities section and I read what I mention in ***). I appreciate your help. Thanks a lot. Regards. Luis Navarro Looking for 700



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06 Jul 2015, 07:53
luisnavarro wrote: Hi Bunuel,
Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because x^2 is negative in the original equation x^22x+15.
*** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<5 and x>3), not IN ranges (5<x<3) .
So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are 5<x<3 (I already studied the inequalities section and I read what I mention in ***).
I appreciate your help.
Thanks a lot.
Regards.
Luis Navarro Looking for 700 My advice would be to brush up fundamentals before answering questions, especially such hard ones. Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmSolving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm
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06 Jul 2015, 13:34
Bunuel wrote: luisnavarro wrote: Hi Bunuel,
Could you explain hoy do you factorize: (x+5)(3−x)>0. ? I have the doubt because x^2 is negative in the original equation x^22x+15.
*** Also I thought that when inequalitie is >0 (Like this exercise), the solution for x are OUT ranges (x<5 and x>3), not IN ranges (5<x<3) .
So I am confused in both things, how do you factorize the inequalitie and also why the solutions for the inequalitie are 5<x<3 (I already studied the inequalities section and I read what I mention in ***).
I appreciate your help.
Thanks a lot.
Regards.
Luis Navarro Looking for 700 My advice would be to brush up fundamentals before answering questions, especially such hard ones. Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmSolving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htmThanks a lot Bunuel, I appreciate your help. I read the information and now it is totally clear to me. Best regards. Luis Navarro Looking for 700.



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Re: M2811
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24 Apr 2017, 00:01
Hi Bunuel, why X<5 and X>3 is not considered?



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16 Jul 2017, 12:13
I have the same question, mkugan80 wrote: Hi Bunuel, why X<5 and X>3 is not considered?



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16 Jul 2017, 21:07
ajanpaul wrote: I have the same question, mkugan80 wrote: Hi Bunuel, why X<5 and X>3 is not considered? Please reread the solution: Quote: Given that \(x\) is an integer from 10 and 10, inclusive (21 values) we need to find the probability that \(x^22x+15\) is greater than zero, so the probability that \(x^22x+15>0\). Factorize: \((x+5)(3x) \gt 0\). This equation holds true for \(5 \gt x \gt 3\). x <  5 and x > 3 is not considered because x's from these ranges does not satisfy \((x+5)(3x) \gt 0\).
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Re: M2811
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23 Oct 2018, 07:28
I tried to factor as written to (x+5)(x3) = m. I found 12 out of 21 possible values for making the left hand side positive and subtracted 12/21 from 1 and got answer E. I think my error was forgetting to subtract out the two cases for making m 0.
Thanks for the problem and explanation.



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Re: M2811
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23 Oct 2018, 08:52
Bunuel wrote: If \(x^2 + 2x 15 = m\), where \(x\) is an integer from 10 and 10, inclusive, what is the probability that \(m\) is greater than zero?
A. \(\frac{2}{7}\) B. \(\frac{1}{3}\) C. \(\frac{7}{20}\) D. \(\frac{2}{5}\) E. \(\frac{3}{7}\) \(x^2 + 2x 15 = m\) i.e. \(x^2 + 2x +1 = 16m\) i.e \(m = 16  (x+1)^2\) For m>0 possible values x can take are 4, 3, 2, 1, 0, 1, 2, 3. Total values x can take 21 (i.e. 10 to 10) Required probability = 7/21 = 1/3
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23 Oct 2018, 14:24
Hi Bunuel, chetan2uI arrived at the correct answer but with a different approach/solution with the idea that we need to count the possible 'm' values but not 'x' values, as you did in OE . Not sure if this one is correct. Let me know if this is correct. Rewriting the given solution, (x+5)(x3)=m Now, substituting values of x from 10 to 10 gives m values as 65,48,33,20,9,0,7,12,15,16,15,12,7,0,9,20,33,36,65,84,105 respectively. Out of these, if we remove repetitions, Count of +ve values of m is 4 (7,12,15,16) Count of all m values is 12 So probability : 4/12 => 1/3If we don't remove repetitions, we get 7/21=>1/3



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14 Aug 2019, 18:47
perfect!










