Who wants 10pts for answering a math equation?

Question

You roll two fair number cubes at the same time. What is the probability that the product of the two number cubes will be a multiple of 3?
A. 5/9
B. 1/3
C. 1/4
D. 1/6

Answer correctly and explain. Best answer gets 10pts:)

This is not my Homework.
I'm just studying for a big test.
Just bored to death too.
Tryin' to give someone on here 10pts.
So don't be so snoby about it.

Answer 1

For the product of the 2 numbers to be a multiple of 3, 1 die needs to show either a 3 or a 6 which is 2 of the 6 numbers.

The probability that this does not happen is

2/3*2/3=4/9

so the probability that the product is a multiple of 3 is

1-4/9 = 5/9

Answer 2

List al the ordered pairs when two dice are rolled:-
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Find outcomes such that products are multiples of 3.
ie 13 ,16 , 23 , 26-----65, 66
There are 20 such outcomes.
Probability = 20 / 36 = 5 / 9
Answer A

Answer 3

For the product of the 2 numbers to be a multiple of 3, 1 die needs to show either a 3 or a 6 which is 2 of the 6 numbers.

The probability that this does not happen is

2/3*2/3=4/9

so the probability that the product is a multiple of 3 is

1-4/9 = 5/9

Answer 4

Possibilities are: 1 and 3, 1 and 6,
2 and 3, 2 and 6
3 and 1, 3 and 2, 3 and 3 ,3 and 4, 3 and 5, 3 and 6 ,
4 and 3, 4 and 6,
5 and 3, 5 and 6,
6 and 1, 6 and 2, 6 and 3, 6 and 4, 6 and 5, 6 and 6
Tis gives 20 favorable results out of 36 possible results, so probability is 20/36 = 5/9.

Answer 5

when 2-dies are thrown there are 21 possible ways when 12 is same as 21 (i.e.
11
12 22
13 23 33
14 24 34 44
15 25 35 45 55
16 26 36 46 56 66)
from these ther are 7-cases for which product of 2 digits that appear on dies
is a multiple of 3.these 7-no.s are 13,23,33,34,35,36,46,56,66.
therefore required probablity is 7/21=1/3

Answer 6

Order doesn't count in this problem (1,2 and 2,1 are the same roll) so
there are a total of 21 possible rolls (count 'em 1,1,1,2,1,3,1,4,1,5;1,6;2,2;etc or use 6!)
7 of these have a sum of 3, 6, 9, or 12 (multiples of 3)
so your probablity is 7/21 simplified to 1/3

Answer 7

you're thoroughly happening the right technique. It seems kinda confusing provided that it would not look such as you will have the skill to factor something rapidly (that would help to sparkling up for x). yet a extra gadget you will have the skill to apply to sparkling up for x is thru employing the easy Theorem of Algebra. extremely, you will have the skill to locate the inspiration in a function that has an n degree of a minimum of one million. To hit upon such root, you look on the main coefficient, in this situation a million (because of the actuality it extremely is x^3), and additionally on the coefficient with n degree of 0, your stable, in this situation -12. next you checklist the climate of each and every particularly one in all them a million: +- a million ----we are in a position to have the skill to call a million q -12: +-a million, +-2, +-3, +-4, +-6, +-12 ----we are in a position to have the skill to call -12 p next you plug in the plausible solutions p/q into the function x^3 + x^2 - 12 , and hit upon which one =0. observe your plausible solutions are p/q = +-a million, +-2, +-3, +-4, +-6, +-12 in case you plug them in, you 'll hit upon that 2 is your respond.

Answer 8

ummm this is your like homework no doubt
do it urself.