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NAND stands for 'Not And', which means the output will send a signal so long as both inputs aren't on. (If both inputs A and B are 1, then the output is 0. All other combinations give an output of 1)
e, e, e, c, g, e, c, g, e, b, b, b, c, g, eb, c, g, e
Possible, yes. Probable, no. If you have tests and quizzes and do all your home work and class work, and depending on your grading scale, if you got a 100 on everything, maybe. It depends on how many tests you've taken before and after that "F". For instance, let's say an A=5, B=4, C=3, D=2, and F=1. If you got that F on just ONE previous test, then you have to get the 1(F=1 remember) up to a 4(B=4). If you have one F and one A you have a total of six(1+5)Now divide that by 2(number of tests)and you get 3, not a B.(4) Now take another test and get an A, and you have 5+5+1=11(A+A+F) Now divide that by 3(number of tests) 3.67 is close to a B(4). So let's try another test and you get an A. Now we have three A's and an F.(5+5+5+1) Now add that up and divide by 4(4 tests) and there's your B. So unless you take four tests and get A's, you won't get that B.That's assuming you got your F on only ONE test. Otherwise, things change again.
(All notes listed are according to treble clef) Whistling: (All notes are natural) B, D, E, D Bass: (Also all notes are natural) G, G, B, B, C, C, D Vocal: B, B flat, B, C, B, B flat, B, D
Doubling on all. B d a b d e b d a b d e all i got sorry
My understanding is that the blood type and Rh factor are independently inherited, so that all the combinations are possible: B+, B-, AB+, and AB-
20
21
There are 16C6 combinations = 16!/(6!(16-6)!) = 16!/(6!10!) = 8,008 possible combinations. Assuming the 16 numbers are the hexadecimal digits 0-F, they are: {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 6}, {0, 1, 2, 3, 4, 7}, ...., {0, 1, 2, 3, 4, E}, {0, 1, 2, 3, 4, F}, {0, 1, 2, 3, 5, 6}, ..., {0, 1, 2, 3, 5, F}, {0, 1, 2, 3, 6, 7}, ...., {0, 1, 2, 3, 6, F}, ...., {0, B, C, D, E, F}, {1, 2, 3, 4, 5}, ..., {9, A, B, C, D, E}, {9, A, B, C, D, F}, {9, A, B, C, E, F}, {9, A, B, D, E, F}, {9, A, C, D, E, F}, {9, B, C, D, E, F}, {A, B, C, D, E, F} I'll let you fill in the missing 7,990 possible combinations.
To find the probability that when rolling a die and tossing a coin, your will obtain an odd on the die OR a heads on the coin, use the addition rule, which is: P(A) + P(B) - P(A and B) = P(A or B In this example, event A is tossing heads on the coin, and event B is rolling odd on the die. What you are trying to solve is actually A U B (A union B) First the sample set of all 12 possible combinations: S={H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} The 6 heads possible combinations are: A={H1, H2, H3, H4, H5, H6| The 6 odd number possible combinations are: B={H1, H3, H5, T1, T3, T5} The 3 combinations these sets have in common, A intersect B: A n B= {H1. H3, H5} There are 12 possible combinations and 6 of those include a heads on the coin. This is 6/12. There are 12 possible combinations and 6 of those include an odd on the die. This is 6/12. There are 12 possible combinations and 3 of those include both an odd on the die, and a heads on the coin. This is 3/12. 6/12 + 6/12 - 3/12 = 9/12 Simplify the above number to 3/4, which is the correct answer to this question. If you draw a Venn diagram, you will see that the set consisting of tails and evens {T2, T4, T6} falls outside the circles. The diagram makes it easy to see that 9 of the 12 possible combinations fall inside the circle, and 3 of the 12 fall outside. Hope this helps someone. I solidified the information for myself by writing it!
Say you have the letters A,B, and C. Here are all the possible combinations. * ABC * ACB * BAC * BCA * CAB * CBA So, 6 if you don't repeat any of the letters. If you DO repeat letters, then simply take the number of letters you have, (3 for instance), and multiply it to the power of the number of letters you have. So, for 3 letters, the formula would be 33 . Or if you had 4 letters it would be 44 and so on.
If the question concerned the number of combinations of three different coins, the answer is 23-1 = 7. If the coins are a,b,and c, the combinations are a, b, c, ab, ac, bc, abc. If two of the coins are the same there are only 5 combinations and if all three are the same there are 3.
Carbon can form up to four bonds due to its ability to share electrons with other atoms.
Scrambling of allelic combinations refers to the random assortment of different versions of genes (alleles) during the process of meiosis, resulting in new combinations of alleles in the offspring. This process increases genetic diversity within a population.
There are 20 unique 5-card hands. Solution: First break apart the player card combinations. You can either use player card 1, player card 2, or both player cards 1 and 2 with the hand. Therefore, you are taking 5 cards, 3 at a time, when playing with both player cards, and 5 cards, 4 at a time, when playing with 1 player card. Mathematically, this can be expressed using the Binomial coefficient 2 x ( 5 choose 4 ) + ( 5 choose 3 ) = Number of possible 5-card hands. 5 choose 4 = 5 : This represents the combinations between 1 player card and 5 table cards, 4 at a time. (http://www.wolframalpha.com/input/?i=5+choose+4) 5 choose 3 = 10 : This represents the combinations between 2 player cards, and 5 table cards, 3 at a time. (http://www.wolframalpha.com/input/?i=5+choose+3) Thus, 2 x ( 5 ) + ( 10 ) = 20 . Further Information: All possible hand combinations are listed below. A and B are the two player cards. 0 through 5 are the five table cards. ----- A 1 2 3 4 A 2 3 4 5 A 3 4 5 1 A 4 5 1 2 A 5 1 2 3 B 1 2 3 4 B 2 3 4 5 B 3 4 5 1 B 4 5 1 2 B 5 1 2 3 A B 2 3 4 A B 3 4 5 A B 1 2 3 A B 4 5 1 A B 5 1 2 A B 4 5 2 A B 4 1 2 A B 5 1 3 A B 5 2 3 A B 1 3 5
There are infinitely many possible solutions. Select ANY number, A, greater than 7.5 and let B = 15 - A Then B is different from A and A + B = A + 15 - A = 15 Since there are infinitely many possible values for A, there are infinitely many solutions.
In a permutation problem such as this, there are six spaces, each of which could be one of two letters. This means there are sixty-four possible combinations. The two obvious combinations are aanwmn and ccpyop. Looking at those, the word is then canyon.