An ordinary (fair) die is a cube with the numbers through on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events. Event A: The sum is greater than 7.
Event B: The sum is divisible by 4 or 6 (or both).

a. 5/12

b. 7/18

Step-by-step explanation:

In this question, we are interested in computing the probabilities.

Firstly, we need to get the sample space of the results. This refer to all possible results that can emerge from the rolling of the two die.

This can be found in the attachment.

What is important to know however is that the total number of possible results is 36;

so let’s compute;

a. For a, let’s circle all sums which are greater than 7 ;

We can see a total of 16 circled sums;

So the probability is 15/36 = 5/12

b. Sum divisible by 4 or 6 or both

Let’s make a circle again ;

So we have a total of 14

So the probability will be 14/36 = 7/18

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

22 and 22

step-by-step explanation:

10 plus 12 equals 22.

12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.