Roger will flip a coin 3 times. Which 2 events are disjoint? A. Getting exactly 1 head and Getting at least 1 tail
B. Getting at least 1 head and Getting at least 1 tail
C. Getting exactly 1 head and Getting at least 2 tails
D. Getting at least 2 heads and Getting at least 2 tails

D. Getting at least 2 heads and Getting at least 2 tails

Explanation:

Choice A is not disjoint because "getting at least 1 tail" could have the sequence TTH. We see that we have two tails and exactly one head. So the events "getting exactly 1 head" and "Getting at least 1 tail" are possible to occur at the same time; therefore, they aren't disjoint events. Disjoint events are two events that cannot occur simultaneously. An example would be flipping heads and tails on the same coin at the same time.

Choice B is also not disjoint. We could have the sequence THT, which has at least one head and at least one tail. "At least" means that amount or more.

Choice C is also not disjoint. We could have the sequence HTT. This has exactly one head and at least two tails.

Choice D is the only thing left. It must be the answer. It is not possible to get 2 heads and 2 tails when Roger only flips the coin 3 times. He would need to flip the coin at least 4 times for this to happen. The portions "at least" don't even need to be considered. So this shows how choice D is disjoint.

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)