Terry has number cube with sides labeled 1 through 6. he rolls the number cube twice. what is the probability that the sum of the two rolls is a prime number or at least one of the rolls is a 3? explain your reasoning

Sum of the points ot the two rolls
          
                 Roll 1           1    2    3    4    5    6
Roll 2 
1                                    2   3    4    5    6    7

2                                    3    4    5    6   7    8

3                                    4    5    6    7    8    9

4                                    5    6    7    8    9    10

5                                    6    7    8    9    10   11

6                                    7    8    9    10   11  12

Prime results:
2: once, 1
3: twice, 2
5: four times, 4
7: six times, 6
11: twice, 2

The easiest way is to count all the results in which none of the rolls was a three and the sum of the two rolls was not a prime number: it is 14 results.

Now subtract those 14 results from the 36 possible outcomes, the difference is the number of results in which either the sum of the rolls is a prime or at least one of the numbers is three: 36 – 14 = 22

Probability = positive events / total events

Probability: 22/36 = 11/18

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: