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

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this