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The total number of possible hands is the number of ways to choose 9 cards out of 36. This can be calculated using the combination formula:

C(36, 9) = 36! / (9! * (36-9)!) = 9,075,135

To calculate the probability of getting all four 1s, we need to count the number of hands that contain all four 1s. There are 4 ways to choose which suit the first 1 comes from, then 3 ways to choose which suit the second 1 comes from (since we can't choose the same suit twice), and so on. So the number of hands with all four 1s is:

4 * 3 * 2 * 1 * C(32, 5)

The factor of 4 * 3 * 2 * 1 comes from choosing which suits the four 1s come from. After those four cards are chosen, we need to choose 5 more cards from the remaining 32. This can be done in C(32, 5) ways.

So the probability of getting all four 1s is:

(4 * 3 * 2 * 1 * C(32, 5)) / C(36, 9) = 48/935.

1  -  (2160 / 3000) =28% discount on the bike.

This is a 24% discount.

The expression simplifies to \(\dfrac{\sqrt[3]{12} + \sqrt[3]{18} + \sqrt[3]{36}}{2}\)

Therefore, the answer is 12 + 18 + 36 + 2 = 68.

We can use the principle of inclusion-exclusion to find the number of students who play both basketball and baseball. The total number of students who play at least one sport is the sum of those who play basketball and those who play baseball, minus the number who play both sports, since they were counted twice.

So we have:

Total = Basketball + Baseball - Both + Neither 26 = 18 + 7 - Both + 5

Simplifying this equation, we get:

Both = 4

Therefore, there are 4 students who play both basketball and baseball.

The probability of choosing one of these students randomly from the class is:

P(both) = number of students who play both sports / total number of students P(both) = 4 / 26 P(both) = 0.1538 (rounded to four decimal places)

So the probability that a student chosen randomly from the class plays both basketball and baseball is approximately 15.38%

Find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9

[u(2+u)+v(-3-v) ]^9

\(\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\ \binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\\)

To get u^2   n must be 1

\((-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\ =(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\ =9v^8\;[u(u+2)\;][(v+3)^{8}]\\~\\ \)

The constant term for  (v+3)^8 = 3^8

So the u^2v^8 term will be in here

\(9v^8\;[u(u+2)\;]3^8\\~\\ 9v^8\;[u^2+2u\;]3^8\\~\\ The \;\;u^2v^8 \text{ term will be}\\~\\ 9*3^8*u^2v^8 = 59049\;u^2v^8\)

LaTex

\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\
\binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\
(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\
=(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\

9v^8\;[u(u+2)\;]3^8\\~\\
9v^8\;[u^2+2u\;]3^8\\~\\
The \;\;u^2v^8 \text{ term will be}\\~\\
9*3^8*u^2v^8 = 59049\;u^2v^8

114th term   (2u - 3v + u^2 - v^2)^9== - 577368 u ^2 v^9

Mouth-watering crunch and juicy fried chicken bursting with Louisiana flavor. Explore our menu, offers, and earn rewards on delivery or digital orders.

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100,000 - 999,999 : 112,496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

Hello,

To find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9, we need to use the binomial theorem to expand the expression and then identify the term that contains u^2 v^9.

The binomial theorem states that the expansion of (a + b)^n can be written as:

(a + b)^n = ∑(k=0 to n) [n choose k] a^(n-k) b^k

where [n choose k] is the binomial coefficient, given by:

[n choose k] = n! / (k! (n-k)!)

Using this formula, we can expand (2u - 3v + u^2 - v^2)^9 as:

(2u - 3v + u^2 - v^2)^9 = ∑(k=0 to 9) [9 choose k] (2u)^(9-k) (-3v)^k (u^2)^{9-k} (-v^2)^k

= ∑(k=0 to 9) [9 choose k] 2^(9-k) (-3)^k u^(18-2k) v^k u^(2k) (-1)^k v^(2k)

= ∑(k=0 to 9) [9 choose k] 2^(9-k) (-3)^k (-1)^k u^18 v^k

= [9 choose 9] 2^0 (-3)^9 (-1)^9 u^18 v^9 + [9 choose 7] 2^2 (-3)^7 (-1)^7 u^14 v^7 + ...

We are interested in the coefficient of u^2 v^9, which appears only in the first term of the expansion. Therefore, the coefficient is given by:

[9 choose 9] 2^0 (-3)^9 (-1)^9 = 1 * (-19683) * (-1) = 19683

Make use of the identity

\(\sin 2A=2 \sin A \cos A.\)

100 000    to   999 999   is   900 000 numbers altogether

6 numbers that add to 8

The sum will be teween  8 and   54

so the sum could be   8,16,24,32,40,48,

800 000

710 000

620 000

611 000

530 000

521 000

511 100

440 000

432 000

431 100

422 000

421 100

411 110

332 000

331 100

322 100

321 110

311 111

222 200

222 110

221 111

There is an awflu lot to count. I guess there is a better method

970 000

961 000

952 000

951 100

943 000

942 100

941 110

933 100

932 110

931 111

922 210

922 111

...

-----------------------

The sum of  the 6 digits must be between 1 and 54  

6 of those are divisable by 8

If they were all equally likely outcomes then the answer would be  6/54. Trouble is they are not all equally likely outcomes.

there are 900 000 sums altogether.   Ifthey were equally likely it would be   6/54 * 900 000 = 100 000  but they are not equally likely.

Yea, I don't know.   

There are three pairs of siblings from different families to be seated in two rows of three chairs in such a way that siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling.

To solve this problem, we can break it down into two cases:

Case 1: The three pairs of siblings are seated in the same row
In this case, there are 3 ways to choose which row they will sit in, and 3! ways to arrange them within that row. However, since no child may sit directly in front of their sibling, once the first pair is seated, there are only 2 possible chairs for the second child of the second pair and only 1 possible chair for the second child of the third pair. Therefore, the total number of arrangements in this case is:

3 x 3! x 2 x 1 x 1 = 36

Case 2: The three pairs of siblings are seated in different rows
In this case, there are 3 ways to choose which pair will sit in the first row, and 2 ways to choose which pair will sit in the second row. Once the pairs are chosen, there are 2! ways to arrange them within each row. Again, since no child may sit directly in front of their sibling, there are 2 possible chairs for each child in the first row and only 1 possible chair for each child in the second row. Therefore, the total number of arrangements in this case is:

3 x 2 x 2! x 2! x 2 x 2 x 1 x 1 = 96

Finally, we can add the two cases together to get the total number of arrangements:

36 + 96 = 132

Therefore, there are 132 ways to seat three pairs of siblings from different families in two rows of three chairs, if siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling. 

is u in awesome math?

a) Use sticks and stones. (20+5-1) choose (5-1) = 24 choose 4 = 10626.

b) If everybody gets one, then we only have to distribute 15 stickers, the same way. (15+5-1) choose (5-1) or 19 choose 4 = 3876. 

Hope it helps!

stars and bars

15C7 = 6435 possible outcomes

Give Dhruv her 2 chos,

now you have  6 friends and 7 chocolates

12C6 = 924 favourable outcomes

924/6435 =  28 / 195    =approx  14% chance

that crazy thanks a lot for this formula.

To count the number of rectangles that can be formed by a 10x10 square and a 7x8 rectangle, we can use the formula:

Number of rectangles = (number of horizontal lines choose 2) x (number of vertical lines choose 2)

For the 10x10 square, there are 11 horizontal lines and 11 vertical lines, so we get:

Number of rectangles in the 10x10 square = (11 choose 2) x (11 choose 2) = 55 x 55 = 3,025

For the 7x8 rectangle, there are 8 horizontal lines and 9 vertical lines, so we get:

Number of rectangles in the 7x8 rectangle = (8 choose 2) x (9 choose 2) = 28 x 36 = 1,008

Now, we can add the number of rectangles in the 10x10 square and the 7x8 rectangle to get the total number of rectangles:

Total number of rectangles = 3,025 + 1,008 = 4,033

Therefore, there are 4,033 rectangles that can be formed by a 10x10 square and a 7x8 rectangle.

nevermind i got the answer of 3025 and 1008 confirm if possible

after 1 steps answer is just 4 because there are 4 directions the fly can move

Let's use the three ice cream flavors A, B, and C. We want exactly three children to choose the same flavor. This means that the other three children must choose different flavors, otherwise, there would be more than three children with the same flavor.

We have three cases to consider:

1. Three children choose flavor A, and the other three children choose one scoop each of flavors B and C.

2. Three children choose flavor B, and the other three children choose one scoop each of flavors A and C.

3. Three children choose flavor C, and the other three children choose one scoop each of flavors A and B.

For each case, we need to determine the number of ways to assign the ice cream flavors to the six children.

Case 1:

- Choose 3 children out of 6 to receive flavor A: C(6, 3) = 6! / (3! * (6-3)!) = 20 ways.
- The other three children must receive one scoop each of flavors B and C. There are 2 ways to arrange this (BCB or CBC).

So there are 20 * 2 = 40 arrangements for Case 1.

Case 2:

- Choose 3 children out of 6 to receive flavor B: C(6, 3) = 20 ways.
- The other three children must receive one scoop each of flavors A and C. There are 2 ways to arrange this (ACA or CAC).

So there are 20 * 2 = 40 arrangements for Case 2.

Case 3:

- Choose 3 children out of 6 to receive flavor C: C(6, 3) = 20 ways.
- The other three children must receive one scoop each of flavors A and B. There are 2 ways to arrange this (ABA or BAB).

So there are 20 * 2 = 40 arrangements for Case 3.

Now, we add up the arrangements from all three cases:

Total arrangements = 40 + 40 + 40 = 120

Therefore, there are 120 ways for the six children to choose ice cream flavors so that exactly one flavor is selected by three children.

UPDATE: My answer "This is what I have so far: 5pi/18, 11pi/18, 17pi/18, 23pi/18, 29pi/18, 35pi/18"

is actually correct no need to answer :).

Reading from the graph, f(-2) = 5/8.