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Helen [10]
3 years ago
7

Explain why there are either two, one, or no solutions for absolute value equatins. Demonstrate an example for each

Mathematics
2 answers:
AleksAgata [21]3 years ago
4 0

<u>No solution</u>: because the absolute value is positive and can't equal a negative

|x| = -1          

<u>One solution: </u>because zero only has one sign

|x| = 0      ⇒ x = 0      

<u>Two solutions:</u> absolute value is positive whether x is positive or negative.

|x| = 2     ⇒ x = 2 or x = -2      

AVprozaik [17]3 years ago
4 0
Absolute value equations |a|=b have two solutions only if b>0, because it means that a=b or a=-b. Example:
|x + 1| = 1 \\ x = 0 \: or \: x = - 2
Absolute value equations |a|=b have one solution only if b=0, because 0=-0. Example:
|x + 1| = 0 \\ x = - 1
Absolute value equations |a|=b have no solution only if b<0, because absolute value can't have a negative value. Example:
|x + 1| = - 1 \\ x \in \varnothing
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E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36And the variance is given by:

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case we have the following distribution given:

X          3      4       5        6

P(X)   0.07  0.4  0.25  0.28

We can calculate the mean with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 3*0.07 +4*0.4 +5*0.25 +6*0.28= 4.74

In order to find the variance we need to calculate first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 3^2*0.07 +4^2*0.4 +5^2*0.25 +6^2*0.28= 23.36

And the variance is given by:

Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924

And the deviation would be:

Sd(X) =\sqrt{0.8924} =0.9447

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