Jolly Goodfellow

Did the 200 not answer at all, or did they answer 'we have NO recovery plan'? Or am I overthinking? Ugh! I'll go with D.

Any chance someone could email me this material. I'd love to get up to the 100 post mark but I won't have the time. My exam is in a month. mrtonybradshaw@gmail.com

If it asks for 3-digit numbers between 100 and 200, why did you include: 2-0-1 2-2-2 2-4-3 2-6-4 2-8-5 Shouldn't the answer be 9? 111 123 132 135 153 147 174 159 195

There are 4 senior managers & 6 junior managers. How many different 3 member groups can formed in which at least one member in group is a senior manager? 1. 48 2. 100 3. 120 4. 288 5. 600

There are a total of 5C2 unique apple pairs that can be selected: 10 total 4 Good Apples :: 1 Bad Apple There are 4C2 ways to pick two good apples: 6 ways Therefore, P(one good apple) = 1 - P(2 good apples) = 1 - 6/10 = 4/10 = 2/5

They ask for the maximum possible length of the shortest piece.

1 can of concentrate to 3 cans of water = 1:3 1/4 can of concentrate to make orange juice Question asks for 200 6oz. servings of orange juice, or 200*6 = 1200oz. total of orange juice. Set up the proportion to find the amount of concentrate is in the 1200 total. 1 can of concentrate is to the total cans for orange juice (4) as x cans of concentrate is to 1200oz. total of orange juice… 1/4 = x/1200 x = 300oz. of concentrate Question asked for total of 12oz. cans of concentrate. 300 / 12 = 25 cans

6 marbles taken from 12 * 5 marbles left taken from 11 remaining marbles = 6/12 * 5/11 = 1/2 * 5/11 = 5/22

Choose the 6 cards other than the blue from 8 total: 6C2 The total amount of combinations: 8C2 6C2 / 8C2 = 15/28 ANOTHER APPROACH: 1st draw: 2/8 you select blue, 1 – 2/8 = 6/8 will not be blue take one out. 2nd draw: 2/7 you select blue, 1 – 2/7 = 5/7 will not be blue 6/8 * 5/7 = 3/4*5/7 = 15/28

Jaune Y = (4 choose 1)*(5 choose 2) = 4*10 = 40 Jaune X = (4 choose 2)*(5 choose 1) + (4 choose 3) = 6*5 + 4 = 34 Total combinations = 9 choose 3 = 84 Probability of Jaune = (40 + 34)/84 = 37/42

Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit) Ways for second couple = 2*4! These cases include an extra case of both couples sitting together Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways) Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4! Total ways to arrange the 5 ppl = 5! Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5 Thus prob of none seated together = 1 - 3/5 = 2/5

(x^2 + y^ 2)/5 = ( (x+y)^2 + (x-y)^2 )/10 How did you go from dividing by 5 to 10? ** nevermind ** I understand it.

For question #2: There are 5 different pairs in the name: AJ EE AT JA TA 5 pairs taken 2 at time: 5C2 = 10 You have 10 possible interchanging pairs of letters in the name. But the pair (EE) would make the name unchanged, so you have a 1/10 chance of keeping the name unchanged. The answer is 10%.

For question #1: 8c2 * 6c2 * 4c2 * 2c2 / 4! => 28 * 15 * 6 * 1 / 24 = 105

If an = an-4 for all n greater than 4, then a5 = a1, a6 = a2, a7 = a3, a8 = a4, a9 = a5 = a1, etc. So the sequence is just: 2, -3, 5, -1, │ 2, -3, 5, -1, │ 2, -3, 5, -1 .... infinitely repeating This repeats in blocks of four. The sum of the first four terms is 3. If we add the first 96 terms, we'll just be adding that same block of four terms 24 times; the sum will be 24 * 3 = 72. If we add on the 97th term, which will be 2, we get the final answer: 72 + 2 = 74.

This discussion on terminating/repeating decimals was pulled from another site: Factor the denominator of the fraction (after it is reduced to lowest terms) into a product of powers of prime numbers. If the prime factors are all 2 or 5, then the decimal will terminate. Any other prime number appears as a factor, then the decimal will be repeating. Example: If the denominator is 51200, then the decimal will terminate, because 51200 = 2^11 * 5^2, and the only prime numbers dividing are 2 and 5. If the denominator is 288, the decimal will repeat, since 288 = 2^5 * 3^2, and the prime number 3 appears in the factorization. Take our problem: 2^3 * 3 * 7 = 189 => NO (has factor other than 2 or 5) 2^2 * 7^2 = 196 => NO 3^2 * 5^2 = 225 => NO 2^4 * 3^2 => NO 2^7 = 128 => YES (prime factors are all 2, so it terminates) Thus, the answer is E: 39/128

Re-work (3^x – 1)^3 (3^x – 1)^3 = 3^3x/3^3 = 3^3x/27 Now, let’s take a look at the problem: If 3^6x = 8,100, what is the value of 3^3x/27? Take the square root of 3^6x = 8,100…you then have, 3^3x = 90. If 3^3x = 90, then substitute this value into 3^3x/27. You then have 90/27 = 10/3.

Didn't know that. Here's an example taken from another site explaining absolute values: |10x + 5|= -5 The absolute value is set equal to a negative number. There is no value of x that we can plug in that will be a solution because when we take the absolute value of the left side it will always be positive or zero, NEVER negative.