The probability of the first event with x possible outcomes is equal to x/30. Also, the probability of the second event with y possible outcomes is y/29.
(1) P of 2 digit number, 21/30
P that the number is 4 is 1/29
Multiplying both probabilities will give us 7/290
(2) P that number is 19 is 1/30
P that the number is multiple by 4 is 6/29
The answer is 1/145
(3) P that number is 24 is 1/30
P that number is less than 15 is 14/29
The answer should be 7/435.
(4) P that number 1st number is perfect square is 5/30
P that second number is perfect square is 4/29.
The answer is 2/87.
Answer:
Step-by-step explanation:
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Answer:
Step-by-step explanation:
a+b+c=0, a+b=-c,a+c=-b, b+c=-a
(a+b+c)^3=(a+b+c)^2*(a+b+c)=(a^2+b^2+c^2+2ab+2ac+2bc)*(a+b+c)=
a^3+ab^2+ac^2+2a^2b+2a^2c+2abc+a^2b+b^3+bc^2+2ab^2+2abc+2b^2c+a^2c+b^2c+c^3+2abc+2ac^2+2bc^2=a^3+b^3+c^3+3a^2b+3a^2c+3ac^2+3ab^2+3bc^2+3b^2c+6abc=
a^3+b^3+c^3+3a^2*(b+c)+3c^2(a+b)+3b^2(a+c)+6abc=
a^3+b^3+c^3+3a^2*(-a)+3c^2*(-c)+3b^2*(-b)+6abc=
a^3+b^3+c^3-3a^3-3c^3-3b^3+6abc=
6abc-2a^3-2b^3-2c^3=2(3abc-a^3-b^3-c^3)=
2*[3abc-(a^3+b^3+c^3)]=0
so 3abc-(a^3+b^3+c^3)=0
so a^3+b^3+c^3=3abc
