What sums are equal to 6/12?

Mathematics

2 answers:

nadya68 [22]3 years ago

6 0

Hey there!

"What sums are equal to $\bold{\frac{6}{12}}$ ?"

$\bold{\frac{6}{12}=\frac{1}{2}=0.5 }$

It can't be $\bold{A. \frac{6}{12} + \frac{6}{12}+ \frac{6}{12}+\frac{6}{12}+\frac{6}{12}+\frac{6}{12}}$ because the answer would be $\bold{3}$
It can be $\bold{B.\frac{2}{12} + \frac{1}{12} + \frac{1}{12} +\frac{2}{12}}$
It can also be $\bold{C. \frac{5}{12} + \frac{1}{12}}$
It can also again be $\bold{\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{3}{12}}$
$\boxed{\boxed{\bold{Answers: B,C,D}}}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

VLD [36.1K]3 years ago

4 0

Answer:

A + c

Step-by-step explanation:

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Two number cubes are rolled for two separate events:

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Let us first make the sample space for the event A. The Sample Space will be:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5)

(5,1) (5,2) (5,3) (5,4)

(6,1) (6,2) (6,3)

As can be clearly seen the Sample Space for the event A has 30 elements in it.

Now, it is given in the question that the event A has already occurred and we need to find the probability of the event B occurring given that the event A has already occurred.

Now, as we can see, from the sample space, only 11 events out of the 30 are the events of interest to us. This is shown in bold below:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5)

(5,1) (5,2) (5,3) (5,4)

(6,1) (6,2) (6,3)

Thus, the probability that the event B occurs given that the event A has already occurred is:

$P(B|A)=\frac{11}{30}\approx0.367$