Question

Asked Sep 6, 2019

Suppose n(X)=240n(X)=240 and XX is partitioned into six subsets X1X1, X2X2, X3X3, X4X4, X5X5, and X6X6. All assume that n(X1)=n(X2)=n(X3)n(X1)=n(X2)=n(X3) and n(X4)=n(X5)=n(X6)n(X4)=n(X5)=n(X6). If there are 3 times more elements in X1 as there are in X5, find the following values:

n(X1)

n(X2)

n(X3)

n(X4)

n(X5)

n(X6)

Step 1

The given details are

n(X) = 240

n(X1) = n(X2) = n(X3) =* x*

n(X4) = n(X5) = n(X6) = *y*

n(X1) = 3n(X5), (i.e.,) *x* = 3*y*

Observe the following from the above details.

Step 2

Obtain the value of *x* and *y* ...

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