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1 year ago

8.5b−3.4∙(13a−3.2b)+a

Mathematics

1 answer:

Nesterboy [21]1 year ago

8 0

Answer:

8.5b - 3.4(13a - 3.2b) + a = 19.4b - 43.2a

Step-by-step explanation:

It is a simple mathematical problem with multiple like terms. We can solve it by applying basic mathematical rules of multiplication and addition/subtration.

8.5b - 3.4(13a - 3.2b) + a

= 8.5b - 3.4*13a -3.4*(-3.2b) + a

= 8.5b - 3.4*13a + 3.4*3.2b + a

= 8.5b - 44.2a + 10.88 b + a

Now, only like terms can be added to each other

= (8.5b + 10.9b) + (a - 44.2a)

= 19.4b + (-43.2a)

= 19.4b - 43.2a

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From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one def

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Answer:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

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

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X 0 1 2 3

P(X) 0.92 0.03 0.03 0.02

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500$

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

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300$

The variance is calculated with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075$

And the standard deviation is just the square root of the variance and we got:

$Sd(X) = \sqrt{0.3075}= 0.5545$

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