Gina randomly chooses from three different games to play on her computer: game A, game B, and game C. She generated this table of random combinations to model the situation. The estimated probability that Gina plays at least three games before she plays game a is

The answer is 0.34

Step-by-step explanation:

I'm sorry i don't have an explanation for this. please like this comment is my answer is right

Explanation:

You must find the probability of Gina playing at least three games before playing game A. The phrase “at least three” means “three, four, five, or more.” That means Gina would not play game A in the first two occurrences (n – 1 = 3 – 1 = 2). Search the table for all outcomes that do not have A in the first two occurrences. These occurrences represent the event you’re interested in. Estimate the probability using the formula

.

The answer is 0.34.

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40