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hoa [83]
3 years ago
6

What is the solution set of the equation x + 6 = 2(x + 3) - x?

Mathematics
1 answer:
Svetach [21]3 years ago
6 0

Answer:

<em>x</em> = all real numbers

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Distributive Property

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

x + 6 = 2(x + 3) - x

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Addition Property of Equality] Add <em>x</em> on both sides:                                     2x + 6 = 2(x + 3)
  2. [Distributive Property] Distribute 2:                                                                 2x + 6 = 2x + 6

Here we see that both sides are identical.

∴ our solution is all real numbers.

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Maslowich

Answer:

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

                               X ~ Po (1.3)

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a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

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a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

                       P(X = x ) =\frac{ (lambda)^k e^(^-^l^a^m^b^d^a^)}{k!}\\\\P(X = x )  =\frac{ (1.3)^k e^(^-^1^.^3^)}{k!}

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

                       P(X = 2 )  =\frac{ (1.3)^2 e^(^-^1^.^3^)}{2!}\\\\P(X = 2 )  = 0.23028    

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

                      P(X < 4 )  = P ( X = 0) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )\\\\P(X < 4 )  = \frac{ (1.3)^0 e^(^-^1^.^3^)}{0!}+ \frac{ (1.3)^1 e^(^-^1^.^3^)}{1!}+ \frac{ (1.3)^2 e^(^-^1^.^3^)}{2!} + \frac{ (1.3)^3 e^(^-^1^.^3^)}{3!}\\\\P(X < 4 )  = 0.27253 + 0.35429 + 0.23028 + 0.09979 = 0.95689

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

                   P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

                                               = P ( X ≥ 3 ) / P (X ≥ 2 )

                                               = P ( X > 2 ) / P ( X > 1 )

                                               = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]  

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

                     P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

                                      = 0.8571

                     P ( X ≤ 1 ) = 0.27253 + 0.35429

                                      = 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

                     P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

                                                  = [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

                                                  = 0.38292

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