A system is comprised of 3 components, each of which is either "working", "iffy", or "failed". The system will work if any two components are working or if none of the components are failed. Let W be the event that the system is working. Let A be the event that at least one component is iffy. a). Define carefully a sample space, S, for this experiment. How many elements does it have? b). In terms of your sample space, give the events W, A, and W ∩ A. fa

I already know that there are 27 elements, but don't understand how did they do part b) for events W?

a. 27 elements

b. W= 11 elements

A = 19 elements

W n A = 19 elements

Step-by-step explanation:

(a).

Let W= Working

Let I = Iffy

Let F = Failed

Let SS = { WWW, WWI, WWF, WIW, WII, WIF, WFW, WFI, WFF, IWW, IWI, IWF, ..., FFF}

This represents the total samples of elements that has a 3--lettered spelling of words with letters W,I and F in no particular order.

Each of the elements has 3 * 3 total elements

For the 3 alphabets, there are 3 * (3 * 3) elements

Thus, there are |SS| = 3³ = 27 elements of SS

(Eaach of 3 positions in the spelling can be any of 3 letters).

(b).

The event W is the set W = { WWW, WWI, WIW, IWW, WWF, WFW, FWW, IIW, IWI, WII,III }.

Thus, W has |W| = 11 elements.

The event A is the set A = { III, IIW, IWI, WII, IIF, IFI, FII, IFF, FIF, FFI, IWW, WIW, WWI, IFW, FIW, FWI, IWF, WIF, WFI }.

This, A has |A| = 19.

Since there are 2³ elements having only letters W and F

2³ = 8

Thus, there are 27 −8 = 19 elements of Sample Space that have at least one “I”

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: