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Concentration: Social Entrepreneurship, Social Entrepreneurship
GPA: 3
WE: Consulting (Insurance)
A number is to be selected at random from the set above. What is the
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25 Oct 2010, 07:43
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{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation \((x - 5)(x + 10)(2x - 5) = 0\)?
Re: A number is to be selected at random from the set above. What is the
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Updated on: 01 May 2012, 02:41
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{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x-5)(x+10)(2x-5) = 0?
A. 1/12 B. 1/6 C. 1/4 D. 1/3 E. 1/2
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Originally posted by Stiv on 01 May 2012, 02:30.
Last edited by Bunuel on 01 May 2012, 02:41, edited 1 time in total.
Re: A number is to be selected at random from the set above. What is the
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25 Oct 2010, 07:50
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vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?
A. 1/12 B. 1/6 C. 1/4 D. 1/3 E. 1/2
Solutions of the equation \((x-5)(x+10)(2x-5)=0\) are: \(x=5\), \(x=-10\) and \(x=2.5\).
There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is \(P=\frac{2}{12}=\frac{1}{6}\).
Re: A number is to be selected at random from the set above. What is the
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20 Jan 2018, 16:07
Hi All,
We're asked for the probability of choosing a number (from a given set of 12 numbers) that will "fit" a given equation. That equation is set equal to 0, which makes solving the equation rather easy. Since we're dealing with the PRODUCT of three "terms", if ANY of those terms equals 0, then the product will equal 0.
IF... we select X = -10 from the set, then we have (-15)(0)(-25) = 0, so X = -10 IS a solution. IF... we select X = 0 from the set, then we have (-5)(10)(-5) = 250, which is NOT a solution.
The given equation has 3 solutions: +5, -10 and +5/2. However, only two of those solutions appear in the set (-10 and 2.5). Thus, the probability of selecting a number that fits the equation is 2/12 = 1/6
A number is to be selected at random from the set above. What is the
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14 Jun 2018, 05:05
Bunuel wrote:
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?
A. 1/12 B. 1/6 C. 1/4 D. 1/3 E. 1/2
Solutions of the equation \((x-5)(x+10)(2x-5)=0\) are: \(x=5\), \(x=-10\) and \(x=2.5\).
There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is \(P=\frac{2}{12}=\frac{1}{6}\).
Re: A number is to be selected at random from the set above. What is the
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14 Jun 2018, 06:39
1
dave13 wrote:
Bunuel wrote:
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?
A. 1/12 B. 1/6 C. 1/4 D. 1/3 E. 1/2
Solutions of the equation \((x-5)(x+10)(2x-5)=0\) are: \(x=5\), \(x=-10\) and \(x=2.5\).
There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is \(P=\frac{2}{12}=\frac{1}{6}\).
You have 3 values of x possible and we need to check how many of these values are part of the set. I don't seem to understand why are you trying to form a quadratic equation
Btw, as for your question, the highlighted portion is wrong
\((x-5)(x+10)(2x-5) = 0\) -> \((x^2 - 5x +10x - 50)(2x - 5) = 0\) -> \((x^2 + 5x - 50)(2x - 5) = 0\) The entire equation will need to be multiplied and we will have a cubic equation
Hope this clears your confusion!
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Re: A number is to be selected at random from the set above. What is the
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14 Jun 2018, 09:10
pushpitkc thanks can you please show how do we find 2 roots: -10 and 2,5, i am kind of confused what to do with highlighted part \((x^2 + 5x - 50)(2x - 5) = 0\)
Re: A number is to be selected at random from the set above. What is the
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14 Jun 2018, 10:54
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dave13 wrote:
pushpitkc thanks can you please show how do we find 2 roots: -10 and 2,5, i am kind of confused what to do with highlighted part \((x^2 + 5x - 50)(2x - 5) = 0\)
Re: A number is to be selected at random from the set above. What is the
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17 Jun 2018, 19:36
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation \((x - 5)(x + 10)(2x - 5) = 0\)?
A. 1/12 B. 1/6 C. 1/4 D. 1/3 E. 1/2
According to the given equation, x can equal the following:
x - 5 = 0
x = 5
or
x + 10 = 0
x = -10
or
2x - 5 = 0
2x = 5
x = 2.5
Since -10 and 2.5 are in the set, the probability of selecting one of those solutions is 2/12 = 1/6.
Answer: B
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