I suspect that you are not stating this correctly. Is condition two supposed to be that 5x+40 is less than 90 degrees — or is it that one of the actual acute angles of the triangle is equal to 5x+40.

In order to form a triangle, X has to be greater than six inches long and less than 14 inches long.

If X is greater than or equal to 10 inches, then 5x+40 is 90 degrees or more — which can’t be an acute angle. I suspect that there are only a fininte number of points where an actual acute angle of the triangle is equal to 5x+40 degrees — to the answer to that question is probability zero.

But if it just means that 5x+40 is less than 90 — then any number between six and ten would meet the criteria — so the answer is 4/20 = 0.2

(i) a + b < a million (ii) a >= 2b or b >= 2a i’m going to be sloppy approximately endpoints, because of the fact that they do no longer influence the risk. restoration a. to fulfill (i), b ought to be interior the kind [0, a million-a]. to fulfill (ii), b ought to be interior the kind [0, a/2] or [2a, a million]. actually we ought to compute the intersection of the 1st variety with the union of the 2d 2, which would be completed with the aid of cases, applying [0, a million-a] ? ([0, a/2] U [2a, a million]) = ([0, a million-a] ? [0, a/2]) U ([0, a million-a] ? [2a, a million]). If a <= 2/3, then a million - a >= a million/3 >= a/2, so the 1st 0.5 of the final expression is purely [0, a/2]. If a >= 2/3, then a million – a <= a million/3 <= a/2, so the 1st 0.5 is purely [0, a million-a]. If a <= a million/3, then a million - a >= 2/3 >= 2a, so the 2d 0.5 is purely [2a, a million-a]. If a >= a million/3, then a million – a <= 2/3 <= 2a, so the 2d 0.5 is empty. The expression is then... 0 <= a <= a million/3: [0, a/2] U [2a, a million-a] a million/3 <= a <= 2/3: [0, a/2] U empty 2/3 <= a <= a million: [0, a million-a] U empty word that a/2 < 2a, so those are each and each disjoint. Their lengths are, so as, a/2 + ((a million-a) - 2a) = a/2 - 3a + a million = a million - 5a/2 a/2 a million-a for sure, lengths listed decrease than are opportunities. Integrating those for a from 0 to a million supplies integral_0^(a million/3) of (a million - 5a/2) da = 7/36 integral_(a million/3)^(2/3) of a/2 da = a million/12 integral_(2/3)^a million of (a million-a) da = a million/18 including them up supplies a million/3, as you claimed.