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Let's suppose that there are 100 students at the  school

Then

21 play football

13 play basketball

25 play  either

N ( play  either)  = N(play football)  + N (play basketball)   - N (who play both )

     25            =             21                +          13      -     N (who play both)

     25   -  21     -13     =    - N (who play both)

             -9  =  -  N (who play both )

              9   = N ( who play both )

M =  3C

F = C + 56

So.......

(C) + ( 3C) +( C + 56)  =   676

5C +  56  = 676

5C  =  676  -  56

5C  =  620

C =   620  / 5    =   $124

M =  3 (124) = $ 372

You can see that it can be divided into two rectangular grids which only touch at 1 lattice point. Name that lattice point M.

Note that to walk from P to Q, your path must pass through M.

Number of ways to walk right/up from the bottom left to top right of a rectangular grid = \(\displaystyle\binom{\text{width} +\text{length}}{\text{width}}\).

So, the number of ways to walk right or up from P to M = \(\displaystyle\binom{2 + 3}{2} = 10\)

The number of ways to walk right or up from M to Q = \(\displaystyle\binom{3 + 4}{4} = 35\).

Now, total number of ways = \(10 \times 35 = 350\)

The longest line segment is the main diagonal.

If you want the length of it:

\(\begin{array}{cl} &\text{Length of main diagonal}\\ =& \sqrt{14^2 + 9^2 + 7^2}\\ =& \sqrt{326} \end{array}\)

We have the formula:

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac{\langle\mathbf v, \mathbf w\rangle}{\|\mathbf w\|^2}\mathbf{w}\)

Now, we compute \(\langle\mathbf v, \mathbf w\rangle\) and \(\|\mathbf w\|^2\).

\(\begin{array}{cl} &\langle\mathbf v, \mathbf w\rangle\\ =& 2(2) + 3(-1) + (-1)(0)\\ =& 1\\ &\|\mathbf w\|^2\\ =& 2^2 + (-1)^2 + 0^2\\ =&5 \end{array}\)

Therefore, substituting back into the formula, we have

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}\)

Dozen=12

no.of roses that are red=6

not red=6

Happy birthday melody!!!

20*4% = 20/25 = 4/5ounces

Ans:4/5 ounces

No it is just  a little kid playing make believe.

Is this a english question?

I would say ... It’s six of one and a half-dozen of the other. 

You may want to consult a second source...

GA

--. .-

Thanks for showig us Alan,

I tried to do somthing similar using matrices but got in a big mess.

I didn't think of the obvious things.  :(

Glowlife,  if you are given an answer, maybe from your teacher, then please show us  (with the logic behind it if possible).

I think your answer is too big Max ://

If the beads were just in a circle, where just rotations are the same then the answer woud be 5! = 120

But there are less options than this because of the reflection aspect.

As far as reflection goes, these 3 are the same

So I am thinking tht the answer might be  120/3  =  40

Hi Max, it is great to see you here again !

It is a good start  

Here is how I approached it

\(xyz=720\\ xy+xz+yz=333\\ \sqrt[3]{333}\approx 8.9\)

this means that at least one of the dimensions is 8 or less, and at least one is 9 or more

Factor  720 and you get    

\(720=2^4*3^2*5\)

the factors 8 and smaller are  1,2,3,4,5,6,8   

Try x=8 

If x=8 then  yz=720/8 = 90

8y+8z+90=333   so     8y+8z = 333-90 = 243   

But 243 is not divisable by 8 so 8 is not one ofthe dimensions.

Try x=6     I ran into the same problem

Try x=5      same problem

Try x=4    same problem

Try x=3

If x=3 then  yz= 720/3 = 240

3y+3z+240=333

3y+3z=93  so y+z=31

31=15+16, that sounds promising

If the sides, x,y and z  are  3, 15, and 16  then

(3*15)+(3*16)+(15*16) = 45+48+240 = 333  Bingo

the sides are   3,15 and 16

(961, 1147, 1271, 1333, 1369, 1457, 1517, 1591, 1643, 1681, 1739, 1763, 1829, 1849, 1891, 1927, 1961, 2021, 2077, 2173, 2183, 2201, 2209, 2257, 2263, 2279, 2419, 2449, 2479, 2491, 2501, 2537, 2573, 2623, 2627, 2701, 2747, 2759, 2773, 2809, 2867, 2881, 2911, 2923, 2993, 3007, 3053, 3071, 3127, 3131, 3139, 3149, 3193, 3233, 3239, 3293, 3317, 3337, 3379, 3397, 3403, 3431, 3481, 3503, 3551, 3569, 3589, 3599, 3649, 3713, 3721, 3737, 3763, 3811, 3827, 3869, 3901, 3937, 3953, 3959, 3977, 4033, 4061, 4087, 4141, 4171, 4181, 4183, 4187, 4189, 4223, 4247, 4307, 4309, 4331, 4343, 4387, 4399, 4429, 4453, 4469, 4489, 4559, 4601, 4633, 4661, 4687, 4699, 4717, 4747, 4757, 4819, 4841, 4847, 4859, 4891, 4897, 5029, 5041, 5063, 5069, 5123, 5141, 5143, 5183, 5207, 5251, 5293, 5311, 5329, 5353, 5371, 5429, 5459, 5461, 5561, 5609, 5617, 5633, 5671, 5699, 5723, 5767, 5777, 5891, 5893, 5917, 5959, 5963, 5969, 5977, 5989, 6059, 6077, 6157, 6161, 6241, 6283, 6313, 6319, 6431, 6439, 6497, 6499, 6527, 6533, 6557, 6649, 6667, 6731, 6767, 6887, 6889, 6893, 6901, 6943, 7031, 7081, 7169, 7171, 7261, 7303, 7313, 7367, 7373, 7387, 7493, 7519, 7571, 7597, 7663, 7729, 7739, 7747, 7811, 7921, 7957, 7979, 7991, 8023, 8051, 8083, 8137, 8201, 8249, 8357, 8383, 8453, 8479, 8509, 8549, 8611, 8633, 8777, 8881, 8927, 8989, 9017, 9047, 9167, 9179, 9271, 9301, 9313, 9379, 9409, 9523, 9563, 9701, 9727, 9797, 9869, 9991) = 233 numbers

They all have prime factors that FALL between 30 & 140

Total price before the discount -> $193

He paid -> $105

Suppose the price of toaster -> $x

the toaster was reduced 20%
                 = x * 80/100 = 1/5 = $4x/5

then, price paid of kettle = 4x/5 * 2/5

                                       = $8x/25

according to the clue

               4x/5 + 8x/25 = 105

             (20x + 8x)/25 = 105

                       x = 93.75 [Total price of kettle]

 reduced price of kettle = 4x/5 = 4 * 93.75/5

                                                  = $75

                       Difference = 93.75 - 75

                                        = $18.75

so, the kettle was reduced by $18.75

Let toaster be Paid = x

Kettle Paid = 2/5x = 2x

Total 7x = 105

           x = 15

Kettle Price = 30

and Toaster Price = 75

Before discount Kettle Price 80% x = 30

                                => x = 30/80 * 100

                                     x = $37.5

Price of toaster before reduced = $95.5

reduced Price = 20.5/95.5 * 100

Kettle reduced = 21.4%

Happy Birthday  Melody

Solutions:

The easiest method to solve this (IMHO) is to note there are (6^5) unique arrangements of five (5) dice, and each with 1 to 6 pips. Reserve two of these dice and populate them with (1, 2) and (2, 1). Populate the remaining three spaces with dice populated with 3 to six pips. Divide this by the total population sample.  

\(\large nPr(5,2) \normalsize* \dfrac{4^3}{6^5} = 0.164609\)

Note this is a permutation of a set of 2 from 5. (A combination will only count the sets). This counts both the sets and their order; that is there are 10 set of (1, 2) and 10 sets of (2, 1) for a total of 20 unique sets. With 2 spaces for two of the dice counted and populated with (1, 2) and (2, 1), the 1’s and 2’s are removed from the remaining sample space giving (4^3). The product of (nPr(5,2) * 4^3 = 1280) gives the number of sets of five with a single “1”and a single “2” in each set. Dividing this by 6^5 gives the probability that a roll of five dice will produce one of these sets.

This method works for any selection of fixed dice with any population size.

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

Other solutions:

Changing CPhill’s equation (C (5,2) * (1/6)^2 * (4/6)^3) to a permutation, presents the correct solution probability.  nPr (5,2) * (1/6)^2 * (4/6)^3 = 0.164609. This equation resembles a binomial probability with a missing fraction.  Note that the exponent (2) corresponds to the (r) and the sum of the exponents (5) corresponds to the (n). Why does this work?  IDK... This is a Nancy Drew mystery, and Gingerlock Holms has yet to solve it.

Best guess: The asymmetry of probability is corrected in the permutations. This equation will work for any (n) when (r=2), with corresponding adjustments to the exponents. It does not work when (r>2).

GA

 --. .-

The question is quite clear. There is no ambiguity; it should not be subjected to second-guessing. 

Five 6-sided dice are rolled. What is the probability that exactly one of the dice shows 1, and one of the dice shows 2? (The comma, hanging there like a limp dick in a love-making session, is superfluous. It neither enhances nor alters the meaning of the statement or question.) 

In any case, neither of the above answers gives the correct solution for this question.

CPhill’s solution is not a binomial probability. The probabilities (1/6 and 4/6) do not sum to one (1).

Mr. BB’s ascii slop and atrocious math presentation is a binomial probability, but it is not a solution for this problem. It sure as hell is NOT a solution for a single “1” and at least one “2”.   

See below for solutions and explanations.

GA

--. .-

Problem:  (1/3)(x + 3)²

The anti-derivative is:  (1/3)·2·(x + 3)  =  (2x + 6)/3  (by using the chain rule)

If you multiply out the original problem:  (1/3)(x + 3)²  =  (1/3)(x² + 6x + 9)

which has:  (1/3)·(2x + 6)  =  (2x + 6)/3  as its anti-derivative; the same as before.

Now:  evaluate  (2x + 6)/3  when x = 2

    and evaluate it when x = 1

    and subtract the second answer from the first.