I think it is D but if it is not I’m sorry
Answer:
second option
Step-by-step explanation:
Given the rule
(x, y ) → (x + 3, y - 5)
This means add 3 to the original x- coordinate and subtract 5 from the original y- coordinate, that is
D(4, - 4 ) → D'(4 + 3, - 4 - 5 ) → D'(7, - 9 )
E(5, - 5 ) → E'(5 + 3, - 5 - 5 ) → E'(8, - 10 )
Respuesta: Uno
La suma de la probabilidad de todos los resultados posibles de un experimento aleatorio debe ser igual a uno.
Porque algunos resultados más ocurren en cada sendero y la suma de todas las probabilidades es 100% o uno.
Espero haber ayudado, buena suerte! :)
If they are in the same line yes. If they aren't then its considered non-collinear.
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
