A consumer organization estimates that 35% of the households in a particular community have one television set, 39% have two sets, and 16% have three or more sets. If two independent households are chosen at random, what is the probability that at least one has one or more television sets

90%

Step-by-step explanation:

Probability of households with one TV = 35% or 0.35

Probability of households with two sets = 39% or 0.39

Probability of households with three or more = 16% or 0.16

The probability of a household with no TV set = 0.10 or 10% (1 - 0.35 - 0.39 - 0.16)

Therefore, the probability of a household with at least one or more sets = 90% (35% + 39% + 16%) or 100% - 10%.

Probability is the chance that something may occur out of many events.  To calculate probability, we divide the number of events by the number of possible outcomes.

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

~

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11