two different floor plans are being offered in a new housing development. several prospective buyers from three different age groups were surveyed about which plan they preferred. find the probability that a randomly selected by her preferred plan a or was in the 40-to 49-year-old age group. (The answer is not 0.912)

Explanation:

Add up the values in the plan A column. There are 10+12+16 = 38 people who prefer plan A.

Add up the values in the "40-49" row to find that 16+8 = 24 people are ages 40 to 49.

We have 38+24-16 = 46 people who either prefer plan A, are aged 40-49, or fit both descriptions. I subtracted off 16 because those 16 people were counted twice when adding 38 and 24.

An alternative way to get this value of 46 is to add up everything that is in column1 or row 3 (or both). So that would get 10+12+16+8 = 46.

Now add up everything in the table to find out how many people were surveyed total. That would be 10+7+12+15+16+8 = 68 people overall.

The probability of someone liking plan A, or being age 40-49, or both is 46/68 = 0.6765 approximately. Rounding to 3 decimal places gives 0.677

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

160

step-by-step explanation:

idk

i hope i