If two fair dice are rolled at the same time, what is the probability that of it obtaining two multiples of 3?

2 answers:

6 0

Im assuming this is a non biased 6 sided dice,

multiples of 3 would be 3 and 6 therefore chance of rolling one would be 2/6 = 1/3.

however since the question says two dice at the same time, you have to multiply the probability which would be 1/3 x 1/3 = 1/9

Bingel [31]3 years ago

6 0

How many multiples of three are there in a dice?
6 and 3. So 2.
And there are 6 possible outcomes for each dice, and both dices added there are 12 possible outcomes. This number will be your denominator.
So since there are 2 in each dice, add them to equal 4, and put the 4 as the nominator.
Anser: 4/12.

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1. Find the product. Simplify. 13/14 of 10/9. A. 15/42 B. 5/21 C. 2/7 D. 5/6 2. Andy has 5 pots, and each pot can hold 3/8 pound

Rufina [12.5K]

1) Resultant fraction: $\frac{65}{63}$

2) Total soil: $1\frac{7}{8}$ pounds

3) Result of the product: $\frac{35}{2}$

Step-by-step explanation:

In this problem, we want to find the product between the two fractions

$\frac{13}{14}$

and

$\frac{10}{9}$

In order to find this product, we have to multiply the numerators of each fraction and the denominators of each fraction. We get:

$\frac{13}{14}\cdot \frac{10}{9}=\frac{13\cdot 10}{14\cdot 9}=\frac{130}{126}$

Now we simplify, dividing both numerator and denominator by 2:

$\frac{130/2}{126/2}=\frac{65}{63}$

And the fraction cannot be further simplified.

Here we have:

n = 5 (number of pots that Andy has)

$s=\frac{3}{8}$ (amount of soil (in pounds) that each pot can contain)

In order to find the amount of soil that Andy needs to fill all the pots, we have to multiply the number of pots (n) by the amount of soil that each pot can contain (s).

If we do so, we find:

$t=n \cdot s = 5 \cdot \frac{3}{8}=\frac{5\cdot 3}{8}=\frac{15}{8}$