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a store gathers some demographic information from their customers the following chart summarizes the age-related information they collected one customer is chosen at random for a prize giveaway answer the following using either fractions or decimals rounded to three places so notice the demographic information is ages and these age groups uh don't overlap so we don't have to worry about any individual person being in two different age groups obviously which means any probabilities that we want to combine we can just add together so let's take a look at the first question what is the probability that the customer is at least 20 but no older than 39 so at least 20 that would be 20 or more so let's just highlight that group so at least 20 would be all of these ranges here because the last one is greater than or equal to 60 so that means anyone that's older than 60 the first one is not included because those are uh people of the ages 0 to 19 because they're less than 20. okay now no but they're no older than 39. so i've highlighted a few groups here that represent folks that are older than 39 so they can't be older than 39 so these three i need to remove now the number of people within these groups we have 75 and 94. so the probability that someone is between the ages of 20 to 39 is going to be 75 plus 94 oops 4 that is divided by the total number of customers altogether so i don't have that right now but that's easily uh determined just using the calculator so i'll go ahead and do that first so 80 plus 75 plus 94 plus 98 66 and 99 so we have 512 customers in this survey so the total is 512. so this is going the probability is going to be the number of uh folks that qualify for our event over the total number of people in the survey so this will be 75 plus 94 which is 169 over 512 and it says you can answer with fractions so that's fine as an answer or you can go ahead and turn that into a decimal and we've got 0.330 so that's going to be approximately 0.330 and that's a fine answer as well what is the probability that the customer is either older than 39 or younger than 20. so let me erase my groups here and do some more highlighting so older than 39 if they're older than 39 they could be 40 to 49 50 to 59 and greater than 60. all of these groups would be older than 39. or younger than 20 younger than 20 would just be this first group here so for this remember when we take two events and we join them together with the word or we add those events together so the probability that they are older than 39 or younger than 20. going to be the sum of all these groups so these numbers here plus that number there so this would be 98 plus 66 plus 99 plus 80 all over the total number of folks in the survey so let's see what we come up with 8 66 99 and 80. so that's 343. so 343 over 512 is a perfectly fine answer or you can change it into a decimal make sure you round properly to three decimals so uh the nine here and the fourth place would bring this nine up which would bring the six up so this would be point six seven zero and i guess if i'm being precise there i should make that an approximation okay what is the probability that the customer is at least 30 all right so if they are at least 30 that means they're 30 or more so that would be 30 to 39 40 to 49 50 to 59 greater than 60. these are all groups of individuals that are more than 30. so i need to add these values together here so the probability that someone is at least 30 so greater than or equal to 30 that's 94 plus 98 plus 66 plus 99 all over 512. so let's take a look 357 over 512 which comes out to 0.697 now um if you don't want to add this many numbers together you can take advantage of complements so the sum of these numbers here plus the sum of these numbers here has to add up to 512. since there's only two of these to add together you can use a complement you could say 80 plus 75 and then subtract that from 512. and that's also going to give you the total of the remaining ones uh because if this sum is 155 then 155 plus this amount is 512 and you're just basically solving a small algebraic problem