Students who party before an exam are twice as likely to fail as those who don't party (and presumably study). If 20% of the students partied before the exam, what fraction of the students who failed went partying? unanswered

The fraction of the students who failed to went partying =

Step-by-step explanation:

Let total number of students = 100

No. of students partied are twice the no. of students who not partied.

⇒ No. of students partied = 2 × the no. of students who are not partied

No. of students partied before the exam = 20 % of total students

⇒ No. of students partied before the exam = × 100

⇒ No. of students partied before the exam =  20

No. of students who not partied before the exam =

Thus the fraction of the students who failed to went partying =

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100