Question attached.
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It cannot be done unless you complete that table. You have given no values for .
Sorry for confusion: 0 = 1/27 1 = 6/27 2 = 12/27 3 = 8/27
So for an answer maybe something along the lines... F(x) = 0 if x<0 F(x) = 1/27 if 0<=x<1 F(x) = 6/27 if 1<=x<2 F(x) = 12/27 if 2<=x<3 F(x) = 8/27 if 3<=x
It will look like this: Now you fill in the blanks.
F(x) = 0 if x<0 F(x) = 1/27 if 0<=x<1 F(x) = 6/27 if 1<=x<2 F(x) = 12/27 if 2<=x<3 F(x) = 1 if 3<=x Okay I see I did a mistake... it should be 1, not 8/27 for the 3<=x
No you missed the accumulation.
Ok I get it now. ACCUMULATION.... aka adding. So we get 0+1/27+6/27+12/27=19/27. Is my 2<=x<3 incorrect though?
Originally Posted by DINOCALC09 Is my 2<=x<3 incorrect though? It is wrong also. Add the 'jump'. When you finish must be continuous on the right.
What do you mean by jump? I know what continuous is kinda, but not sure how it relates to inequalities.
Originally Posted by DINOCALC09 What do you mean by jump? It should go from to . That is a jump of .
yes, i see that. however, don't see what's wrong with the inequality necessarily. Should 2 not be inclusive?
Frankly this is tiresome. It is I am done with the thread.
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