geno3141

Could you rewrite the fourth sentence?

This is the part I don't understand:  "When you are of the way back you realize you don’t need the book".

Question #2:  There are 4 red, 5 yellow and 6 blue, for a total of 15.

                      There are  ₁₅C₃  to choose 3 marbles from 15.

To get 2 red, 1 yellow, 0 blue:  ₄C₂ x ₅C₁ x ₆C₀

To get 2 red, 0 yellow, 1 blue:  ₄C₂ x ₅C₀ x ₆C₁ 

To get 1 red, 2 yellow, 0 blue:  ₄C₁ x ₅C₂ x ₆C₀ 

To get 0 red, 2 yellow, 1 blue:  ₄C₀ x ₅C₂ x ₆C₁​ ​ 

To get 1 red, 0 yellow, 2 blue:  ₄C₁ x ₅C₀ x ₆C₂

To get 0 red, 1 yellow, 2 blue:  ₄C₀ x ₅C₁ x ₆C₂

Add all the individual possibilities for red, yellow, and blue, and divide by the total number possible.(​ ​ ₁₅C₃ ).

Question #1:  According to the other answer, there are 720 swappy six-digit numbers; however, I got 810. Where did I go wrong?

My way:  If you start with this list:

     10 xx 10     11 xx 11     12 xx 12     13 xx 13     14 xx 14      ...     97 xx 97     98 xx 98     99 xx 99

There will be 90 in this list and each will be a palindrome if you switch the last two digits.

However, 9 of them are already palindromes:  11 xx 11     22 xx 22     ...     88 xx 88     99 xx 99.

So, removing them, there will be only 81 in the list.

Now, for each of the 81 in the list, there are 10 ways to replace the "xx"; for example:

     47 00 47     47 11 47     47 22 47     ...     47 88 47     47 99 47.

Multiplying 81 x 10, I get 810.

What mistake did I make?

In number 3, can you use a calculator?

If so:                               cos(5°)  =  sin(25°)  + sin(X)

                        cos(5°) - sin(25°)  =  sin(X)

Using inv(sin):

     --->        =  35°             

I'll try #!:

Using the formula:  cos(X) + cos(Y)  =  2·cos( (X + Y)/2 ) ·cos( (X -Y)/2 ):

Let  ax = 4x  and  bx  =  4x:  cos(ax) + cos(bx)  =  cos(4x) + cos(4x)   2 · cos( (4x + 4x)/2 ) · cos( (4x - 4x)/2 )

                                                                           =  2 · cos( 8x/2 ) · cos( 0/2 )

                                                                           =  2 · cos(4x) · cos(0)

                                                                           =  2 · cos(4x) · 1

                                                                           =  2 · cos(4x)

Let  cx = 3x  and  dx  =  1x:  cos(ax) + cos(bx)  =  cos(3x) + cos(x)   2 · cos( (3x + x)/2 ) · cos( (3x - x)/2 )

                                                                          =  2 · cos( 4x/2 ) · cos( 2x/2 )

                                                                          =  2 · cos(2x) · cos(x)

So we have:  2 · cos(4x) · 2 · cos(2x) · cos(x)  =  4 · cos(4x) · cos(2x) · cos(x)

Choose  cos(4x) + cos(4x) + cos(3x) + cos(1x)

Count the total number of possibilities:  36

Count the number which are red and have a number less than 5:  4

Count the number which are yellow and have a number less than 5: 4

Add these together (=8) and divide by the total number of possibilities (=36).

Probability  =  8/36

This is an arithmetic series, first term = 0.30, last term = 0.50, with the number of terms = 21.

A formula for the sum of an arithmetic series is:  Sum  =  [ number · (first + last) ] / 2

                                                                             Sum  =  [ 21 · (0.30 + 0.50) ] / 2

                                                                             Sum  =  8.40

This can be checked by summing all the individual terms.

x + y  =  10     --->     y  =  10 - x

Substitute this value into:  x² + y²  =  56

                                x² + (10 - x)²  =  56

                     x² + (100 - 20x + x²)  =  56

                            2x² - 20x + 100  =  56

                              2x² - 20x + 44  =  0

                               x² - 10x + 22  =  0   

Using the quadratic formula with  a = 1,  b = -10, and  c = 22:

                        x  =  [ 10 +/-  sqrt( 10² - 4(1)(22) ) ] / ( 2·1 )

                        x  =  [ 10 +/- sqrt(12) ] / 2

                        x  =  5 +/- sqrt(3)

Substituting these values back into  y  =  10 - x     --->     y  =  5 -/+ sqrt(3)

Answers:     ( 5 + sqrt(3), 5 - sqrt(3) )   and   ( 5 - sqrt(3),  + sqrt(3) )

There are 99 numbers in the range 1...99, 50 are odd and 49 are even.

For the sum of two numbers to be even, either both are even or both are odd.

We have to remember that a and b must be different numbers.

If a = 9, b can be 49 of the 50 odd numbers.

If a = 18, b can be 48 of the 49 even numbers.

If a = 27, b can be 49 of the 50 odd numbers.

Continue this pattern and add the results.

1)  There are 15 rolls that result in A larger than B:  

     (6,5)    (6,4)    (6,3)    (6,2)    (6,1)    (5,4)    (5,3)    (5,2)    (5,1)    (4,3)    (4,2)    (4,1)    (3,2)    (3,1)    (2,1)

     Probability  =  15/36.

2)  There are 10 rolls that result in a sum of 5 or less:   

      (4,1)    (3,2)    (2,3)    (1,4)    (3,1)    (1,3)    (2,2)    (2,1)    (1,2)    (1.,1)

     Two of these have dice that are equal.

      Probability  =  2/10.

3)  There are only 6 results that have both dice landing on the same number: 

      (1,1)    (2,2)    (3,3)    (4,4)    (5,5)    (6,6)

      Therefore, there are 30 results that have different numbers.

      There are 8 results that have a difference of 2:  (6,4)    (4,6)    (5,3)    (3,5)    (4,2)    (2,4)    (3,1)    (1,3)

      Probability  =  8/30.