Suppose that 43% of all adults in a certain community are obese and that 36% suffer from diabetes. If 20% of the adults in this community are both obese and suffering from diabetes, what is the probability that a randomly chosen adult in this community is obese or suffers from diabetes (or both)? Write your answer as a decimal (not as a percentage)

0.59

Step-by-step explanation:

Since the probability of adults in this community that are both obese and suffering from diabetes overlap, when you add 43% and 36%, you count the percentage that overlap twice (Once the obese adults and once for the diabetic adults). Therefore, you have to subtract 20% from 43% + 36% so you only count the overlap once. Solving, 43% + 36% - 20% = 59%, or 0.59.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier