How are common denominator and a common multiple alike and different

1 answer:

7 0

They both have to do something with common has to find the multiples the difference is sorry only one :( The common multiple has to find the multiple of the denominator of the number Hope this helps

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16. From past experience, a company has found that in cartons of transistors, 92% contain no defective transistors, 3% contain o

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Answer:

E(x)=0.15

V(x)=0.3075

S(x)=0.5545

Step-by-step explanation:

The mean of a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)+...+x_n*p(n)$

where $x_1,x_2,...,x_n$ are the values that the variable can take and $p(x_1),p(x_2),...,p(x_n)$ are their respective probabilities.

So, if we call x the number of defective transistors in cartons, we can calculate the mean E(x) as:

$E(x)=(0*0.92)+(1*0.03)+(2*0.03)+(3*0.02)=0.15$

Because there are 0 defective transistor with a probability of 0.92, 1 defective transistor with a probability of 0.03, 2 defective transistors with a probability of 0.03 and 3 defective transistors with a probability of 0.01.

At the same way, the variance V(x) is calculated as:

$V(x)=E(x^2)-(E(x))^2$