If possible, find all values of a such that there are no x-intercepts for f(x) = 3|x + 4| + a. (enter your answer using interval notation. if an answer does not exist, enter dne.)

We are given function f(x) = 3|x + 4| + a.

And it is said that a should be all those values such that there are no x-intercepts for given function f(x).

We know, the standard absolute function f(x) = a|x-h| have an x-intercept at (h,0).

But if we write fuction in form f(x) = a|x-h|+k, the value of k represents k units up or down of the x-axis.

We could also say, if we have value k other than 0, it will never touch( or cut)  the x-axis.

With respect to the given function the k is take by a variable.

So, the value of a can't be 0.

Therefore, we can write interval notation for a values as  

(-∞,0) ∪ (0,∞).

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )