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%)

answer: the common difference is -0.10

the answer is:

the correct option:

b.

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

then, using placing logarith with base 7 on both sides of the equation, we have:

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

hence, we have that the correct option:

b.

have a nice day!