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Mathematics 20 Online
OpenStudy (anonymous):

Ben wants to play a carnival game that costs $2. In his pocket he has 5 red tickets worth 35 cents each, and 15 blue tickets worth 10 cents each. Which of the following systems of inequalities correctly represents the constraints on the variables in this problem? Let r represent the red tickets and b represent the blue tickets. A. 5r + 15b >/ 2 b / 2 b / 2 0 / 2 0

OpenStudy (anonymous):

@dtan5457 @Godlovesme @Nnesha

OpenStudy (godlovesme):

@Nnesha help plz idk sorry =0 @Tallan

OpenStudy (anonymous):

I just need help to answer this question fairly quick

OpenStudy (anonymous):

@iambatman

OpenStudy (godlovesme):

@Directrix @acxbox22 @WhiteFangIII

OpenStudy (anonymous):

I need all the help i can get on this. :P

OpenStudy (anonymous):

what should be my best guess @Godlovesme

OpenStudy (godlovesme):

hang on i'm asking ppls

OpenStudy (anonymous):

ok

OpenStudy (jdoe0001):

the carnival game costs to play $2, $2 is really 200 cents Ben has, 5 "r" tickets, worth 35c each 5 * 35 = 175 he also has 15 "b" tickets, 10c worth each, 15 * 10 = 150 we just happen to know that he has 5 red ones, or r = 5 and that he has 15 bliue ones, b = 15 what if we didn't know how many "r" and "b" he had? let us say he had "r" red, and "b" blue ones how much worth? well r * 35c = 35r b * 10 = 10b we know that, to play he has to have 200cents at least so, whatever 35r + 10b is, has to be AT LEAST 200 so.. what do you think?

OpenStudy (anonymous):

would it be .35 and .10 or just 35 an 10

OpenStudy (jdoe0001):

if you want to put it in dollar terms recall that there are 100cents in $1 so 35cents is really 35/100 = 0.35 35r in dollar terms would be then 0.35r 10 cents is 10/100 = 0.10 so 10b in dollar terms is 0.10b

OpenStudy (jdoe0001):

using cent terms would be \(\bf 35r+10b\ge 200\) in dollar terms it'd be \(\bf \cfrac{35}{100}r+\cfrac{10}{100}b\ge\cfrac{200}{100}\implies 0.35r+0.10b\ge 2\)

OpenStudy (jdoe0001):

greater than or equals meaning if what Ben has is 200 cents, is ok, he can play the carnival game if he has more than 200, say 250 or 300, is also ok, he'd just be left with some leftover change to buy cotton candy or so if he has less than 200, well, then he couldn't really play it so it has to be 200 or more, or \(\bf \ge 200\)

OpenStudy (anonymous):

ok so can I take a guess at the answer?

OpenStudy (jdoe0001):

well is obviously either B or C B has an issue \(\bf b \le 15\) means that "b" could be say a negative value, like -5, -25 same goes for saying that \(\bf r \le 15\) so, whatever the amount of "b" and "r" tickets are they have to be more than or equal to 0 they both can't be 0 of course, just one or the other if "r" is 0, I assume "b" will take the slack and if "b" is 0, "r" takes the slack

OpenStudy (jdoe0001):

\(\bf r\le 5\) I meant, anyhow so should be obvious which is it from there

OpenStudy (godlovesme):

awesome job @jdoe0001 :)

OpenStudy (anonymous):

So it would be B then @jdoe0001

OpenStudy (anonymous):

?

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