Math algebra?
For these questions you must use the Fundamental Counting Principle. The FCP says (mostly) that to find the total number of possible outcomes, you multiply the possible outcomes at each step. You must also divide out possibilities that look the same. #1. You want to pick three people from 15. For the 1st position, you have 15 people possible to choose from. For the 2nd position you now have only 14 people to choose from (you're not going to pick the 1st person again!). For the 3rd position you have 13 people to choose from. Therefore, the total number of possibilities is: 15x14x13. #2 Think of each letter of the word SEVENTEEN on a separate card. Turn the cards face down on a table and select one card for the 1st letter in your pretend word. The number of possibilities for that 1st letter is any of the 9 cards. For the 2nd letter you have 8 cards to pick from. For the 3rd letter you have 7 cards and so on. So, you get 9x8x7x6x5x4x3x2x1 possibilities. OK so far? But now we need to get rid of the results that are duplicates. We get duplicates because 4 of the cards have E and 2 of the cards have N. Picking one E card for the 2nd letter does not give us a different outcome than if we had picked another E card for the 2nd letter. So what we must do is divide by all of the pretend words that spell the same thing but had the E cards switched among themselves or the N cards switched among themselves. Duplicates made by E cards: use the original word SEVENTEEN. How many E cards could you use for that 2nd letter? 4. Pick it. How many E cards could you now use for that 4th letter? 3. How many E cards could you now use for the 7th letter? 2. And the number of E cards you have left for that 8th letter is 1. Thus, the number of duplicates made by E cards is: 4x3x2x1 Likewise, the number of duplicates created by the N cards: 2x1. So the total different permutations for the letters in the word SEVENTEEN is: (9x8x7x6x5x4x3x2x1)/(4x3x2x1x2x1) = 7,560 We didn't divide the answer to #1 by anything because none of the people were identical (like the E cards or N cards were). Hope this wasn't too confusing! p.s.The shorthand way to write the answer to #2, with factorials, is: 9!/(4!2!)
Math problem help? Algebra 2?
How do you solve this? You buy 15 articles of clothing at a local clothing store. Each shirt cost $3 and each pair of pants costs $10. The total cost is $94. How many shirts and how many pants did you buy?
Algebra 2 math problems?
1) the sum of the digits of a two digit number is 13. if the digits are reversed, the new number is 27 more than the original number. what is the original number? 2) marci has a total of 50 stamps, consisting of 25-cent stamps and 15-cent stamps. if she doubles the number of 25-cent stamps and halves the number of 15-cent stamps, they will be worth $16.50. how many 25-cent stamps and how many 15-cent stamps were there in the original 50? Can you also show me the steps too? thank you.
Algebra 2 math question? Help plz?
i'm having trouble on this math question even though i think it's simple i just cant get it, here it is: As a fund-raiser, the student council sold t-shirts and sweatshirts. They sold a total of 105 t-shirts and sweatshirts and raised $1170. If the cost of a t-shirt was $10 and the cost of a sweatshirt was $15, what was the number of sweatshirts sold? answers are... A.) 24 B.) 52 C.) 81 D.) 105
Pre-algebra math question !?
Easy! :) 124-(6*(0+15)) = 34! You're welcome, haha.
How does one solve an SAT algebraic math problem?
When taking an SAT test, the questions are written in a manner such that if you make the proper observation you can quickly solve the problem in your head.Consider the following problem:If the equation [math]\displaystyle \frac{1}{15} - \frac{7x}{5} = 21[/math] holds true, what is the value of [math]\displaystyle \frac{1}{3}-7x[/math] ?In the case of the problem above, the writer of the question wants you to see that 5 is a factor of both denominators on the left hand side of the equation and can be eliminated by multiplying 5 on both sides of the equation. The result is then 5 x 21 which equals 105.The writer of the question does not want you to solve for x first, even though that is another way to solve the problem.Time is an element of the SAT test. When it comes to mathematics problems, if you solve all of them using mechanical methods (i.e. brute force computation of x first), you will struggle meeting the time allocation for the test.
Math homework help!! Algebra in the let/then format?
No need to get frustrated in advance. 1) A jar contains $36.25 in dimes and quarters. There are 250 coins in the jar. How many quarters are in the jar? You can do this question all in dollars or all in cents. We will use cents because it's easier to use integers than decimals. Let q = number of quarters Then 250 - q = number of dimes Now write the equation that adds up the value of the coins. 25q + 10(250 - q) = 3625 25q + 2500 - 10q = 3625 15q = 3625 - 2500 = 1125 q = 1125/15 = 75 There are 75 quarters. _____________________________ 2) In a pile of coins, there are 15 more quarters than loonies (A loonie is a one dollar coin). The total value of the coins is $21.25. How many quarters are there? Let q = number of quarters Then q - 15 = number of loonies Now write the equation that adds up the value of the coins. 25q + 100(q - 15) = 2125 25q + 100q - 1500 = 2125 125q = 2125 + 1500 = 3625 q = 3625/125 = 29 The number of quarters is 29.
Math Word Problem (algebra) - 10 points?
I keep getting the wrong answer for this and it's making me stabby. Q.1 4 is subtracted from a number and the difference is multiplied by 2. If 15 is added to the product and the sum is divided by 7, the result is equal to 9. Find the number. I am quite sure the first step is to multiply both sides of the equation by 7, giving x - 4 x 2 +15 = 63, no? Or is it {(x - 4) x 2} + 15 = 63? Anyone who can offer a detailed explanation (not just solving it but explaining why as well with words) will get the 10 points. Thanks so much :)