Compute the VC dimension of the following hypothesis sets. (i) H = {h : R → {−1, +1}|h(x) = +1, ∀x ∈ [a, [infinity]), a ∈ R} ∪ {h : R → {−1, +1}|h(x) = +1, ∀x ∈ (−[infinity], a], a ∈ R}. To clarify, the first subset is the positive ray, and the second subset is the negative ray. So the union is the set of all positive rays and negative rays. (ii) H = {h : R → {−1, +1}|h(x) = +1, ∀x ∈ [a, b], a, b ∈ R} ∪ {h : R → {−1, +1}|h(x) = −1, ∀x ∈ [a, b], a, b ∈ R}. So this is the union of the positive intervals and the negative intervals. (iii) H = {h : R 2 → {−1, +1}|h(x) = +1, ∀x where kxk2 ≤ b, b ∈ R}. Note that this is a concentric circle. (iv) H = {h : R 2 → {−1, +1}|h(x) = +1, ∀x where kx − ak2 ≤ b, a ∈ R 2 , b ∈ R}.

answer: h

step-by-step explanation:

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

8/k

step-by-step explanation:

believe it or not anna is correct good job anna give me a fist bump there u go gal