Determine whether of not the following are binomial. Explain why or why not.?
821305
a) effective or not - 2 choices - binomial\nb) smoke or note - 2 choices -binomial\nc) answers could be 1,2,3,4,5,6 . . .therefore NOT binomial
577657
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
390304
THE POSSIBILITIES:\n\n1.1:1→1*1=1\n2.1:2→1*2=2\n3.1:3→1*3=3\n4.1:4→1*4=4\n5.1:5→1*5=5 √\n6.1:6→1*6=6\n\n7.2:1→2*1=2\n8.2:2→2*2=2\n9.2:3→2*3=5\n10.2:4→2*4=8\n11.2:5→2*5=10 √\n12.2:6→2*6=12\n\n13.3:1→3*1=3\n14.3:2→3*2=6\n15.3:3→3*3=9\n16.3:4→3*4=12\n17.3:5→3*5=15 √\n18.3:6→3*6=18\n\n19.4:1→4*1=4\n20.4:2→4*2=8\n21.4:3→4*3=12\n22.4:4→4*4=16\n23.4:5→4*5=20 √\n24.4:6→4*6=24\n\n25.5:1→5*1=5    √\n26.5:2→5*2=10   √ \n27.5:3→5*3=15   √\n28.5:4→5*4=20   √\n29.5:5→5*5=25   √\n30.5:6→5*6=30   √\n\n31.6:1→6*1=6\n32.6:2→6*2=12\n33.6:3→6*3=18\n34.6:4→6*4=24\n35.6:5→6*5=30   √\n36.6:6→6*6=36\n\nProbability = No. of required possibilities/No. of total possible events =11/36\nSo the right answer is D
197973
Yes, that is correct.\n\nThe first round has 2^32 outcomes.  For each of those outcomes, you have 2^16 outcomes.  For each of those you have 2^8, followed by 2^4, 2^2, and finaly 2^1 outcomes.  So multiply that all out and you get 2^63.
89444
The answer is "i" because there is no number, that when mutliplied by itself, will give you -1.  Think about it: the only possible combination that would make sense is 1 X 1, which equals 1...not -1.
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
9363
No.  You can't definitively "prove" a sequence of numbers is not randomly generated by just looking at the numbers, but you can still draw conclusions with a high degree of confidence.\n\nFor example, it is extremely unlikely that a random number generator will produce the same number twenty times in a row, but technically, it is possible.  So seeing this happen doesn't prove it's not random, but you can still conclude that it's not.
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
795924
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
1076304
(1) 1.3\n(4) 4\n(5) Right\n(6) 0\n(7) 0\n(8)0\n(9)12
1090404
look at what is added (in parentheses)\n\n2 (+1) 3 (+2) 5 (+0) 5 (+3) 8 (-1) 7 (+4) 11\n\nStarting with the first and taking every other number you have +1 +0 -1.  The other values are +2 +3 +4.  So you have two patterns interwoven.  Following those patterns will yield:\n\n2, 3, 5, 5, 8, 7, 11, 9, 14, 11, 17, 13, 20
1082362
Sometimes - rarely - there are.  I have very occasionally come to the end of the puzzle with two numbers that could go anywhere in two squares, ie, 2 and 6 might be interchangeable.  Then I think I'm wrong, and will do it all again to be faced with the same outcome!
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
847139
There are clues to help you, but it still requires some guess-and-check- you just get fewer choices.  For example if the middle number is large, like here, you know that the first and last have to be big and the outer and inner have to be small.  You know that if the original terms of the trinomial had no common factors, then the two binomials won't either.  So (2x-4) for example would never be a factor, so don't waste time guessing it.  You can guess the signs (but may have to reverse them) by looking at the signs in the trinomial: both plus means both plus; minus then plus means both minus; plus then minus means one of each; both minus means one of each.
24599
binomial has higest degree of 2 whereas polynomials have more. Polynomials(at least the ones that are going to show up on your math test) can be expressed as product of two or more binomials
880382
Ok, I think that I got it. on the third one you posted, it's about simplifying the binomials (2 numbers). So you add the exponents of the like variables (or coefficiants- there's no number, then it's assumed to be 1. Ex.\n(z^4)^-3 * (z^-2)^-5\n(z^2)^-8\n\nAre there no answers in the back of the book? I would email you, but it blocks your address, if I could see some of your notes it would be easier to explain I think.
2775
They are not inconsistent. They have a solution (x,y)=(0,4). \n\nThey are not dependent, since one is not a multiple of the other. \n\nThey are independent.