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How to solve this math problem??

How to solve this math problem?? Topic: Strategies of problem solving math
June 26, 2019 / By Earlene
Question: 48 students were interviewed about the classes they were taking this semenster: 27 said they were both taking science 18 said they were both taking math 15 said they were both taking english 10 said they were taking both science and math 5 said they were taking both science and english 8 said they were taking both math and english 2 said they were taking all three classes how many students are not taking science, math, or english? how many students are taking exactly one science, math, or english.
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Best Answers: How to solve this math problem??

Celestine Celestine | 3 days ago
That kind of question is answered using the three circles strategy. I think the question is wrong, Please check the number of students that will take math. It cannot be 18 only. Check out what I did. http://cid-001ad034300e3805.skydrive.live.com/self.aspx/Public/3%20circles.pdf
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Celestine Originally Answered: My brother is really good at math, I got him a gift card for Christmas, and I want him to solve a math problem?
Assuming there are unknown amount of oranges in the basket. Firstly, give half of the amount of oranges and another half of orange to person A. Secondly, give half of the rest of the amount of oranges and another half of orange to person B. Thirdly, give half of the rest of the amount of oranges and another half of orange to person C. Same goes to person D, E, F, G and H and eventually there are no oranges left in the basket. Hence, what is the total amount of oranges that person B, E and F will get?
Celestine Originally Answered: My brother is really good at math, I got him a gift card for Christmas, and I want him to solve a math problem?
Determine the value of 1/[sqrt(1)+sqrt(2)] + 1/[sqrt(2)+sqrt(3)] + 1/[sqrt(3)+sqrt(4)] + ... + 1/[sqrt(2600)+sqrt(2601)] The solution is to rationalise the denominator of each term, yielding [sqrt(2) - sqrt(1)] + [sqrt(3) - sqrt(2)] + [sqrt(4) - sqrt(3)] + ... + [sqrt(2601) - sqrt(2600)] = sqrt(2601) - sqrt(1), since all the other terms cancel out... This is equal to 51 - 1 = 50. Haha. Alternatively, you can try this one: There is a deadly disease in town whereby 1 in 100 people will be infected by the disease. A test kit has been developed for this disease. However, it is only 99% accurate, which means that there is a 1% chance that the test result is wrong (turning up positive when you don't have the disease and vice versa). Now, John just took the test and his results turned out to be positive. What is the chance, in %, that John is infected with the disease? The answer, as you desired, is 50%! This might seem unintuitive but it's true if we do some simple calculations: If the results turned out positive, then either (a) John HAS the disease and the test kit was CORRECT, or (b) John DOES NOT have the disease and the test kit was WRONG. Probability that John HAS disease, test kit CORRECT = 1/100 x 99% = 99/10000 Probability that John DOES NOT have disease, test kit WRONG = 99/100 x 1% = 99/10000 Therefore probability that John has the disease given that the result was positive is 99/10000 / (99/10000 + 99/10000) = 50% Bingo.
Celestine Originally Answered: My brother is really good at math, I got him a gift card for Christmas, and I want him to solve a math problem?
Your brother has a gift card for $50 and he wants to spends it all in one particular department store. If the tax rate in the store is 7%., what should the maximum amount of his purchases that he can have so that thei total of $50 will be totally spent?

Annitra Annitra
To me that's a dumb ?. I mean I think it's dumb that teacher's would give you stuff like that because to me it don't make sense. Because you got 48 students, but all the numbers that you have add to more than 48 students so I don't know.
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Annitra Originally Answered: Math Problem! Solve problem using the elimination method?
The elimination method means to take two equations and add them together so that one of your two variables is eliminated. Variables: c = number of children, a = number of adults Now find your two equations from the sentences: '2200 people enter the fair' gives us a + c = 2200 '$5050 is collected' gives us $4.00a + $1.50c = $5050, or 4a + 1.5c = 5050 So our equations are a + c = 2200 4a + 1.5c = 5050 We want to add them together to eliminate one variable, but that won't happen. However we can multiply the first equation by -4, then the a's will be eliminated -4a - 4c = -8800 + 4a + 1.5c = 5050 = -4a + 4a - 4c + 1.5c = -8800 + 5050 -2.5c = -3750 c = -3750/(-2.5) = 1500 children a + c = 2200, so a + 1500 = 2200, so a = 2200 - 1500 = 700 adults
Annitra Originally Answered: Math Problem! Solve problem using the elimination method?
so which you recognize the fringe is all factors extra jointly. as a results of fact it extremely is a rectangle you recognize that the two factors in the time of from one yet another are the comparable length. So assuming the dimensions = x, you recognize that width = a million/2x+2. So upload up your 4 factors, x + x + a million/2x+2 + a million/2x + 2 to equivalent your perimeter of 40 8. x + x + (a million/2x + 2) + (a million/2x + 2) = 40 8 remedy for x. 3x + 4 = 40 8 3x = 40 4 (by ability of subtracting 4 from the two facet of equation) x = 14 2/3 (by ability of dividing the two facet by ability of three) replace your x answer of length back right into a million/2(14 2/3) + 2 for width a million/2 of 14 2/3 + 2 = 9 a million/3 that's proper i'm fairly particular. length is 14 2/3 and width is 9 a million/3

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