"the average starting salary of individuals with a master's degree in statistics is normally distributed with a mean of $48,000 and a standard deviation of $6,000. what is the probability that a randomly selected individual with a master's in statistics will get a starting salary of at least $55,000?

12.1%

Step-by-step explanation:

First calculate the z-score:

z = (x − μ) / σ

z = (55000 − 48000) / 6000

z = 1.17

Look up in a z-score table.

P(z>1.17) = 1 − 0.8790

P(z>1.17) = 0.1210

There's a 12.1% probability that a randomly selected individual with a master's in statistics will have a starting salary of at least $55,000.

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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