At a highschool students can choose between 3 art electives and 4 history electives and 5 computer electives Each student can choose two electives . What is the approximate probability that a student uses a computer elective and an art elective

The answer to the question is

The approximate probability that a student uses a computer elective and an art elective is 15/144 ≈ 0.104 or 10.4 %

Step-by-step explanation:

The available options are

3 art electives, 4 history electives, 5 computer electives

Therefore the total number of electives = 3+4+5 =12

The probability of choosing a computer elective P(computer) = 5/12

and the probability of choosing an art elective P(art)= 3/12

Therefore the combined probability of selecting an computer and an art elective is 3/12×5/12 = 15/144 =0.104 or 10 %

probability for j is 1/4

while

a is 1/6

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 ☺