Question

I will mark brainliest for correct answers!!

Answer 1

Answer:

8%

Step-by-step explanation:

Hello,

8% of the students take neither History or French

so we have 8*200/100=8*2=16 students  out of French and History

let s say that

a is the number of students taking only History

b is the number of students taking both History and French

c is the number of students taking only French

10% of those who take History also take French

so 0.10(a+b)=b <=> 0.10a+0.10b=b

<=> 0.10a+0.10b-0.10b=b-0.10b=0.9b

<=> 0.10a=0.90b

let's multiply by 10 it comes a = 9b

4 times as many students take History as take French

so a + b = 4 (b + c)

it comes 9b + b = 10b = 4b + 4c

<=> 10b-4b=4b+4c-4b=4c

<=> 6b=4c

<=> 3b=2c

<=> c = 3b/2

and we know that a + b + c = 200 - 16 = 184

so

9b + b + 3b/2 = 184 we can multiply by 2 it comes

20 b + 3b = 184*2

23b = 184*2 = 23 * 8 *2 = 23*16

b = 23*16/23 = 16

so b = 16

c = 3*16/2 = 24

c = 24

a = 9b = 144

a = 144

you can see the Venn diagram below

and then the probability that a student picked at random does History and French is 16/200 = 8%

so the answer is 8%

hope this helps