An integer n between 1 and 99, inclusive, is to be chosen at random. what is the probability that n(n + 1) will be divisible by 3 ?

You want to multiply two consecutive numbers, and see when you get a multiple of three.

Since two consecutive numbers share no divisors (except 1), either n or n+1 must be a multiple of 3.

So, your experiments succeeds if n is itself a muliple of three, or if lies immediately before one. So, 2 numbers out of 3 are ok.

Let me try to visualize the situation: I'll list some integers. Below, I'll write if they are multiple of three (M) or not (N). Above, I will write if they will cause the experiment to succeed (S), i.e. n(n+1) will be dibisible by 3, or to fail (F), i.e. n(n+1) will not be a multiple of three. We have

So, since there are exactly 99 numbers between 1 and 99, and 2/3 of the numbers work, the probability that you will choose a number n such that n(n+1) is divisible by 3 is 2/3.

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant

can i see a photo of the question so i can you?