Aconsumer organization estimates that 32% of the households in a particular community have one television set, 38% have two sets, and 16% have three or more sets. if two households are chosen at random, what is the probability that at least one has one or more television sets

If two households are chosen at random, the probability that at least one has one or more television sets is equal to 0.9804 or (98.04%)

Step-by-step explanation:

Then 100% - ( 32%+ 38% + 16% ) = 14% of households does not have any television set

the probability that both of the two randomly chosen households does not have any television set is 0.14×0.14 =0.0196

If two households are chosen at random, the probability that at least one has one or more television sets is equal to

1 - (the probability that both of the two randomly chosen households does not have any television set) = 1- 0.0196 = 0.9804 (98.04%)

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: