A local anime fan club surveyed its members regarding their viewing habits last weekend, and the following information was obtained: 35 members had watched an episode of Naruto, 48 had watched an episode of Death Note, 24 had watched both an episode of Naruto and an episode of Death Note, and 12 had watched neither Naruto nor Death Note. (Round your answers to three decimal places.) (a) What percent of the club members had watched Naruto or Death Note?
(b) What percent of the club members had watched only Naruto?
(c) What percent of the club members had watched only Death Note?

a) 83.099% of the club members had watched Naruto or Death Note.

b) 15.493% of the club members had watched only Naruto

c) 33.803% of the club members had watched only Death Note.

Step-by-step explanation:

First we find the number of students which each habit, treating the habits as Venn sets.

I am going to say that:

Set A: Watched an episode of Naruto.

Set B: Watched an episode of Death Note.

24 had watched both an episode of Naruto and an episode of Death Note:

This means that

48 had watched an episode of Death Note

This means that .

We have that:

In which B - A are those who watched only Death Note. So

35 members had watched an episode of Naruto

This means that .

At least one:

This is

So

Total number of students:

59 at least one, 12 none, so 71 total.

(a) What percent of the club members had watched Naruto or Death Note?

At least one, so 59 out of 71

59*100/71 = 83.099

83.099% of the club members had watched Naruto or Death Note.

(b) What percent of the club members had watched only Naruto?

11 out of 71. So

11*100/71 = 15.493

15.493% of the club members had watched only Naruto.

(c) What percent of the club members had watched only Death Note?

24 out of 71. So

24*100/71 = 33.803

33.803% of the club members had watched only Death Note.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.