As the saying goes, “You can't please everyone.” Studies have shown that in a large
population approximately 4.5% of the population will be displeased, regardless of the
situation. If a random sample of 25 people are selected from such a population, what is the
probability that at least two will be displeased?
A) 0.045
B) 0.311
C) 0.373
D) 0.627
E) 0.689

Step-by-step explanation:

The correct answer is (B).

Let X = the number of people that are displeased in a random sample of 25 people selected from a population of which 4.5% will be displeased regardless of the situation. Then X is a binomial random variable with n = 25 and p = 0.045.

P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – binomcdf(n: 25, p: 0.045, x-value: 1) = 0.311.

P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – 0C25(0.045)0(1 – 0.045)25 – 25C1(0.045)1(1 – 0.045)24 = 0.311.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.

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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