Suppose that we have a sample space S = {E1, E2, E3, E4, E5, E6, E7},
The following probability assignments apply: P(E1) = 0.15, P(E2) = 0.20, P(E3) = 0.10, P(E4) = 0.25, P(E5) = 0.20, P(E6) = 0.05, and P(E7) = 0.05.
Let
A = {E1, E4, E6}
B = {E2, E4, E7}
C = {E2, E3, E5, E7}.
find
A ∪ B =
P(A ∪ B) =
P(A ∩ B) =
Bc =
P(Bc) =

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

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.

just sum

10 + 10 = 20