Whats The Answer Marking Brainliest, Please explain your D...

() (98 points)to be fair, you have to have a very high iq to understant the jokes in rick and morty. the humour is extremely subtle, and without a solid grasp of theoretical physics most of the jokes will go over a typical viewer's head. there's also rick's nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from narodnaya volya literature, for instance. the fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they're not just funny- they say something deep about life. as a consequence people who dislike rick & morty truly are idiots- of course they wouldn't appreciate, for instance, the humour in rick's existential catchphrase "wubba lubba dub dub," which itself is a cryptic reference to turgenev's russian epic fathers and sons. i'm smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as dan harmon's genius wit unfolds itself on their television screens. what fools.. how i pity them. and yes, by the way, i do have a rick & morty tattoo. and no, you cannot see it. it's for the ladies' eyes only- and even then they have to demonstrate that they're within 5 iq points of my own (preferably lower) beforehand. nothin personnel kid

(a) (8%) compute the probability of an even integer among the 100 integers 1! , 2! , 3! , until 100! (here n! is n factorial or n*(n-1)*(n-2) *… 1) (b) (16%) compute the probability of an even integer among the 100 integers: 1, 1+2, 1+2+3, 1+2+3+4, …., 1+2+3+… + 99, and 1+2+3+… + 100

Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [−1, 1] no, f is continuous on [−1, 1] but not differentiable on (−1, 1). no, f is not continuous on [−1, 1]. yes, f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.