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We are on number 11. Sonay customize her bicycle by exchanging the front wheel for a wheel that had ½ the diameter of the back wheel. So this will be a weird looking bicycle. You have a small wheel in front and then let’s see if I can draw a bicycle, then you have a huge wheel of the back because the diameter of this wheel is half of the diameter of that wheel.
Now, when Sonay raised the bicycle, how many revolutions does the front wheel make for each revolution of the back wheel? When you do this problem what you needed, so how many revolution will the front will make for each in the entire back wheel and then let me fix you right it said says that the front wheel has half the diameter of the back wheel.
So, let me to ask you question, if the back wheel goes around, well that’s essentially way of just to asking, let’s make up numbers, let’s say that the back wheels, the back wheels diameter or let’s say it’s radius, this is the back wheel has a radius of two. So, then what’s it’s circumference, it’s circumference for the back wheel would be 2∏R so it would be two times pi times two which is equal to 4∏.
This circumference if this has half the diameter so it will also be half the radius, so the circumference would just be 2∏R which it should be 2∏. So, this one – if you have half the diameter you also have half the circumference. And so when this wheel does one revolution, when this wheel does one revolution, how far did the bicycle traveled assuming the wheel hasn’t scale or anything? Well, when this back does one revolution the bicycle would’ve traveled 4∏ whatever our units, 4∏ feet if that’s our unit.
So, the bicycle travels 4∏ unit it’s going to travel 4∏ units of the front of bicycle as well. So, if we travel 4∏ units how many times this wheel is have turned around to travel 4∏ units. Well, in one revolution this wheel its 2∏ so this wheel actually has to turn around twice and you get turns around twice for everyone revolution of the big wheel. Hopefully that makes sense, you just realize at one revolution of the big wheels is going to be circumference of the front and so the front has to turn around twice when the bike goes over that same distance.
So, the answer to their question is how many revolutions will the front wheel make for each revolution of the back wheel? Well, the answer is C, woops that looks like an A but the answer is C, two.
Problem number 12: A list of numbers contains P+ and N - numbers, so P positive N negative, if a number is pick at random from the list the probability that the number is positive is 3/5s what is the value of N/P? So, the probability of getting a positive is N, is no, no is 3/5s. So what is the probability of getting a positive, a probability of getting a positive is going to be equal to P/N or it’s going to be the probability of the number of positive numbers we have over the total numbers we have, that’s the probability of getting a positive number.
If this was red marbles and blue marbles the probability getting a red marble would be in another number of red over the total number of marbles.
So similar here, it’s the probability it’s the number of positive numbers over the total numbers are the positive numbers plus the negative numbers. And now they are asking as for what is the value -- well again I just want to make sure I’m doing right, if the numbers pick in random so probably that the number is positive, this is probably this positive, it’s 3/5, what is the value of N/P, so now we’ll solve N/P.
So, we could cross multiply here, so we could say 5P, 5P is equal to three time this is equal to 3P plus 3N, I just cross multiplied and then subtract P from both sides you get 2P is equal 3N and we want to figure out N/P. So, let’s divide both sides by P, so you get two is equal to 3N/P and let’s divide both sides by three and you get 2/3 is equal to N/P and we are done, N/P is equal to 2/3 and that’s choice C.
Problem number 13: Clear image and our color is and I will do this in this tacky green, problem 13. The total daily cost C in $ of producing X units of a certain product is given by the function, cost function is C(x) is equal to 600X - 200/X plus K and they are telling us K s a constant and X is less than equal to 100X is less than our equal to a 100. If 20 units were produce yesterday cost of $640 what is the value of K? So units, so they’re essentially saying that C(20) the cost they’re producing 20 units is equal to $640.
So we can just use this information to now solve fro our case. So C(20), so C(20) is going to be equal, put in 20 to this equation so it’s 600 times 20 minus 200, all of that over 20, C is 20th all over 20. And we also —plus K, I forget to plus K, plus K is equal to, that is equal to 640. And what is this equal to?
But we actually don’t haven’t to multiply because this is 600 times 20 and what is 200, 200 is ten times 20, so you can divide the top and the bottom by 20, so we get 20, 20, 20, so you’re just left with 600 and you could multiply that out if know what I just did. But you get 600 - 10 + K = 640, so you get 590 + K is equal to 640, subtract 590 from both sides you get K is equal to 50. And that is choice B. so, you just have to set, figure out, substitute the X in for intercourse unction and then set that equal to $640 and you will solve for K.
Problem number 14: For how many ordered pairs are positive integers XY? positive integers, integers XY is 2X 3Y < 6, so these are positive integers. So, I mean we could figure them out, the zero, we can’t use zero, so if X is, XY, if X is one what can equals? So then we have two, then we have two plus 3Y < 6, this is a case of X is equal to one, so then we have 3Y < 4, so the only, so we have Y would have to be less than four thirds. So the only situation or Y is less than 4/3rd is Y would have to be one because that’s way positive integer is less than 4/3rd.
If X is two then we have four plus 3Y, two times two is < six and so you have 3Y < 2, Y < 2/3, well I don’t see any value where, I don’t see any positive integers than less than 2/3, so X can’t even be equal to two. so, neither of this numbers can get any bigger than one or one, so there’s only one ordered pair positive integers that positive integers and that’s the trick that satisfy this equation one and one. And we know because when we just raise one of them above two, we get above six no matter what. We can’t find the positive integer that satisfies it. So, that’s it. It’s just choice A, not too bad and I will see in the next video.
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