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Mathematics 9 Online
OpenStudy (anonymous):

i need help with word problems

OpenStudy (lgbasallote):

i give help for word problems :P as long as i know it lol

OpenStudy (anonymous):

ok can you help i need to get better at those for the ASVAB

OpenStudy (anonymous):

I also give help for word problems.

OpenStudy (anonymous):

But we need to see the question.

OpenStudy (anonymous):

I don't know what kind of word problems you need help with, so no I can't.

OpenStudy (anonymous):

4. Of 150 people polled, 105 said they rode the city bus at least three times per week. How many people out of 100,000 could be expected to ride the city bus at least three times each week?

OpenStudy (anonymous):

plz help

OpenStudy (anonymous):

So for that question can you tell that you'll have to take the percentage of people who said they did and apply it to the second number? So what would you do first?

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

The question is telling you that the percentage of people who replied yes, it can be assumed the same percentage would apply yes regardless of how many people where polled. So first you need to find out what that percentage is.

OpenStudy (anonymous):

idk how

OpenStudy (anonymous):

Okay, to find a percentage of something you think of it this way, for example: One out of every 2. If you set that up in an equation it would be 1/2 or half or .5 or 50% (Percent is /100, so to find percent you would multiply by 100, to take from percent down into a decimal you would divide by 100). If one out of every 2 people say the ride the bus three times a week, that's 50 percent of them. So apply that to the numbers in the problem you gave, what would it look like then?

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