X. 1. 2. 3. 4. 5. 6 are all the possible events when rolling a die. If it is a fair die then the probability of rolling any number is 1/6. So the probability of rolling at most a two is p(1)+ p(2)= 2/6 or 1/3
X^2+3x+3+[(2x^2+13+15)/(x+5)] is equivalent to option 4. im sorry if i got my math wrong but i hope this helps.
She is incorrect because 27 is not a prime number but yet it has 7 at the end
Answer:
d=107
Step-by-step explanation:
7+5^2 *4
PEMDAS
Since there are no parentheses, we do exponents next
7+25*4
Then we multiply
7+100
Then we add
107
