In a study, 44% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 14 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 14 adults are randomly selected, 3 or fewer are in excellent health. Group of answer choices

Step-by-step explanation:

Probability(P) (k events out of n trials) = nCk * p^k * (1-p)^(n-k), where p=0.40, n=10 and nCk is the number of combinations of n things taken k at a time

:

P ( k < or = 3 ) = P ( k = 3 ) + P ( k = 2 ) + P ( k = 1 ) + P ( k = 0 )

:

P ( k = 3 ) = 10C3 * (0.40)^3 * (0.60)^(10-3) = 0.2149

:

P ( k = 2 ) = 10C2 * (0.40)^2 * (0.60)^(10-2) = 0.1209  

:

P ( k = 1 ) = 10C1 * (0.40)^1 * (0.60)^(10-1) = 0.0403

:

P ( k = 0 ) = 10C0 * (0.40)^0 * (0.60)^(10-0) = 0.0060

:

P ( k < or = 3 ) = 0.2149 + 0.1209 + 0.0403 + 0.0060 = 0.3821 is approximately 0.38

justin bieber ml.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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