Afghanistan, Antigua and Barbuda, Australia, Azerbaijan, The Bahamas, Bosnia and Herzegovina, Canada, Central African Republic, Guatemala, Jamaica, Madagascar, Malaysia, Mauritania, Nicaragua, Panama, Paraguay, Saudi Arabia, Trinidad and Tobago, United Arab Emirates
not grammatically correct, but mathematically correct
The principal quantum number of the first d subshell is 3. In the case of d orbitals, they start appearing in the n=3 energy level.
Here's a simple way to solve it: 2*3=6 6*3=18 18*3=54 54*3=162 162*3=486 486*5=2,430 2,430*7=17,010 In simpler terms, 2*3*3*3*3*3*5*7=17,010.
You have 2 fours so you would write 42 There are 5 threes that follow, so you would write 35 all together you would have: 42 x 35
3 times 63360 is 190,080.
The word "banana" consists of 6 letters, with the letter "a" appearing 3 times, "n" appearing 2 times, and "b" appearing once. To find the number of unique arrangements, you can use the formula for permutations of multiset: ( \frac{n!}{n_1! \times n_2! \times n_3!} ), where ( n ) is the total number of letters, and ( n_1, n_2, n_3 ) are the counts of each distinct letter. This gives ( \frac{6!}{3! \times 2! \times 1!} = 60 ) unique arrangements of the letters in "banana."
Canada or Panama.
The word "DECAGON" has 7 letters, with the letter "A" appearing once, "C" appearing once, "D" appearing once, "E" appearing once, "G" appearing once, "N" appearing once, and "O" appearing once. To find the number of different 4-letter permutations, we need to consider combinations of these letters. Since all letters are unique, the number of 4-letter permutations is calculated using the formula for permutations of n distinct objects taken r at a time: ( P(n, r) = \frac{n!}{(n-r)!} ). Here, ( n = 7 ) and ( r = 4 ), so the number of permutations is ( P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840 ). Thus, there are 840 different 4-letter permutations that can be formed from the letters in "DECAGON."
1/12 possibilty.
The word "CANADA" consists of 6 letters, with the letters A appearing 3 times, and the letters C, N, and D appearing once each. To find the number of distinguishable 6-letter words, we can use the formula for permutations of multiset: [ \frac{n!}{n_1! \times n_2! \times n_3! \ldots} ] Here, ( n = 6 ) (total letters), ( n_1 = 3 ) (for A), and ( n_2 = n_3 = n_4 = 1 ) (for C, N, and D). Thus, the number of distinguishable words is: [ \frac{6!}{3! \times 1! \times 1! \times 1!} = \frac{720}{6} = 120. ] So, there are 120 distinguishable 6-letter words that can be formed from "CANADA."
The first country to win the world cup 3 times was Brazil
"NO". (Norway's 3-letter country code is "NOR").
The chance of any 1 number appearing in a die roll 3 times in a row is: 1/6 * 1/6 * 1/6 which is 1/216 = .00463
3 times
Powers are how many times you multiply a number by itself. Three squared (or three to the second power) is the same as 3 times 3. Three to the third power is the same as 3 times 3 times 3. This can continue forever. For the record, three to the third power is equal to 27.
no
John Edward Cross Country - 2006 Remember the Good Times 3-3 was released on: USA: 2008